Quadratic Equations

A quadratic equation is an equation of the form:

$a{x}^2+bx+c=0$

Where:

• $a,b,c$ are constants, and $a\ne 0$.
• $x$ is the variable we are solving for.

Quadratic equations are fundamental in algebra and appear frequently in various mathematical problems, including geometry, physics, engineering, and economics. The solutions to quadratic equations are called the “roots” of the equation.

Forms of Quadratic Equations

1. Standard Form:
   The most commonly used form of a quadratic equation is the standard form:
   $a{x}^2+bx+c=0$.
2. Factored Form:
   A quadratic equation can sometimes be factored as:
   $a(x–r_1)(x–r_2)=0$
   where $r_1$ and $r_2$ are the roots of the equation.
3. Vertex Form:
   Another form is the vertex form:
   $a(x–h{)}^2+k=0$
   where $(h,k)$ is the vertex of the parabola.

Methods to Solve Quadratic Equations

There are four main methods for solving quadratic equations:

1. Factoring
2. Completing the Square
3. Quadratic Formula
4. Graphing

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1. Solving Quadratic Equations by Factoring

This method is useful when the quadratic equation can be factored easily. To solve by factoring, we write the quadratic equation in its factored form and then set each factor equal to zero.

Example 1: Solve $x^2+5x+6=0$ by factoring.

$x^2+5x+6=0$

First, factor the quadratic:
$(x+2)(x+3)=0$

Set each factor equal to zero:
$x+2=0$ or $x+3=0$

Solve for $x$:
$x=-2$ or $x=-3$.

Thus, the roots are $x=-2$ and $x=-3$.

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Example 2: Solve $2x^2–4x–6=0$ by factoring.

$2x^2–4x–6=0$

First, factor out the GCF:
$2(x^2–2x–3)=0$

Then factor the quadratic:
$2(x–3)(x+1)=0$

Set each factor equal to zero:
$x–3=0$ or $x+1=0$

Solve for $x$:
$x=3$ or $x=-1$.

Thus, the roots are $x=3$ and $x=-1$.

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2. Solving Quadratic Equations by Completing the Square

Completing the square involves manipulating the equation so that one side becomes a perfect square trinomial. This method works for any quadratic equation, but is especially useful when factoring is not possible.

Example 3: Solve $x^2+6x–7=0$ by completing the square.

Step 1: Move the constant term to the other side:
$x^2+6x=7$

Step 2: Take half of the coefficient of $x$, square it, and add it to both sides:
Half of $6$ is $3$, and $3^2=9$.

$x^2+6x+9=7+9$

Step 3: Write the left side as a square and simplify the right side:
$(x+3{)}^2=16$

Step 4: Take the square root of both sides:
$x+3=\pm 4$

Step 5: Solve for $x$:
$x=-3\pm 4$

Thus, $x=1$ or $x=-7$.

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Example 4: Solve $2x^2+8x–10=0$ by completing the square.

Step 1: Divide through by 2 to make the coefficient of $x^2$ equal to 1:
$x^2+4x=5$

Step 2: Take half of the coefficient of $x$, square it, and add it to both sides:
Half of $4$ is $2$, and $2^2=4$.

$x^2+4x+4=5+4$

Step 3: Write the left side as a square and simplify the right side:
$(x+2{)}^2=9$

Step 4: Take the square root of both sides:
$x+2=\pm 3$

Step 5: Solve for $x$:
$x=-2\pm 3$

Thus, $x=1$ or $x=-5$.

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3. Solving Quadratic Equations Using the Quadratic Formula

The quadratic formula can solve any quadratic equation and is derived from completing the square. The quadratic formula is:

$x=\frac{-b\pm \sqrt{b^2–4ac}}{2a}$

Where $a,b,$ and $c$ are the coefficients from the quadratic equation $a{x}^2+bx+c=0$.

Example 5: Solve $x^2+6x+8=0$ using the quadratic formula.

Step 1: Identify $a=1$, $b=6$, and $c=8$.

Step 2: Use the quadratic formula:
$x=\frac{-6\pm \sqrt{6^2–4(1)(8)}}{2(1)}$

$x=\frac{-6\pm \sqrt{36–32}}{2}$

$x=\frac{-6\pm \sqrt{4}}{2}$

$x=\frac{-6\pm 2}{2}$

Step 3: Solve for $x$:
$x=\frac{-6+2}{2}=-2$
$x=\frac{-6–2}{2}=-4$

Thus, the roots are $x=-2$ and $x=-4$.

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Example 6: Solve $2x^2–3x+1=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=-3$, and $c=1$.

Step 2: Use the quadratic formula:
$x=\frac{-(-3)\pm \sqrt{(-3{)}^2–4(2)(1)}}{2(2)}$

$x=\frac{3\pm \sqrt{9–8}}{4}$

$x=\frac{3\pm \sqrt{1}}{4}$

$x=\frac{3\pm 1}{4}$

Step 3: Solve for $x$:
$x=\frac{3+1}{4}=1$
$x=\frac{3–1}{4}=\frac{1}{2}$

Thus, the roots are $x=1$ and $x=\frac{1}{2}$.

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4. Solving Quadratic Equations by Graphing

Quadratic equations can also be solved by graphing the corresponding quadratic function $y=a{x}^2+bx+c$. The solutions are the points where the graph intersects the $x$-axis.

Example 7: Solve $x^2–4x–5=0$ by graphing.

Step 1: Rewrite the equation as $y=x^2–4x–5$.

Step 2: Graph the quadratic function.
The graph is a parabola that opens upward, with the vertex at $(2,-9)$.

Step 3: Identify the points where the graph intersects the $x$-axis.
The graph intersects the $x$-axis at $x=-1$ and $x=5$.

Thus, the roots are $x=-1$ and $x=5$.

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Examples 8 to 100

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Example 8: Solve $x^2+3x–10=0$ using the quadratic formula.

Step 1: Identify $a=1$, $b=3$, and $c=-10$.

Step 2: Use the quadratic formula:
$x=\frac{-3\pm \sqrt{3^2–4(1)(-10)}}{2(1)}$

$x=\frac{-3\pm \sqrt{9+40}}{2}$

$x=\frac{-3\pm \sqrt{49}}{2}$

$x=\frac{-3\pm 7}{2}$

Step 3: Solve for $x$:
$x=\frac{-3+7}{2}=2$
$x=\frac{-3–7}{2}=-5$

Thus, the roots are $x=2$ and $x=-5$.

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Example 9: Solve $3x^2–7x+2=0$ using the quadratic formula.

Step 1: Identify $a=3$, $b=-7$, and $c=2$.

Step 2: Use the quadratic formula:
$x=\frac{-(-7)\pm \sqrt{(-7{)}^2–4(3)(2)}}{2(3)}$

$x=\frac{7\pm \sqrt{49–24}}{6}$

$x=\frac{7\pm \sqrt{25}}{6}$

$x=\frac{7\pm 5}{6}$

Step 3: Solve for $x$:
$x=\frac{7+5}{6}=2$
$x=\frac{7–5}{6}=\frac{1}{3}$

Thus, the roots are $x=2$ and $x=\frac{1}{3}$.

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Example 10: Solve $x^2–10x+25=0$ by factoring.

Solution:
Recognize that this is a perfect square trinomial:
$x^2–10x+25=(x–5{)}^2$

Thus, the root is $x=5$ (with multiplicity 2).

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Example 11: Solve $2x^2–5x+3=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=-5$, and $c=3$.

Step 2: Use the quadratic formula:
$x=\frac{-(-5)\pm \sqrt{(-5{)}^2–4(2)(3)}}{2(2)}$

$x=\frac{5\pm \sqrt{25–24}}{4}$

$x=\frac{5\pm \sqrt{1}}{4}$

$x=\frac{5\pm 1}{4}$

Step 3: Solve for $x$:
$x=\frac{5+1}{4}=\frac{6}{4}=1.5$
$x=\frac{5–1}{4}=\frac{4}{4}=1$

Thus, the roots are $x=1.5$ and $x=1$.

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Example 12: Solve $x^2+7x+12=0$ by factoring.

Solution:
Find two numbers that multiply to 12 and add to 7. These numbers are 3 and 4.

$x^2+7x+12=(x+3)(x+4)$

Thus, the roots are $x=-3$ and $x=-4$.

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Example 13: Solve $5x^2–6x+1=0$ using the quadratic formula.

Step 1: Identify $a=5$, $b=-6$, and $c=1$.

Step 2: Use the quadratic formula:
$x=\frac{-(-6)\pm \sqrt{(-6{)}^2–4(5)(1)}}{2(5)}$

$x=\frac{6\pm \sqrt{36–20}}{10}$

$x=\frac{6\pm \sqrt{16}}{10}$

$x=\frac{6\pm 4}{10}$

Step 3: Solve for $x$:
$x=\frac{6+4}{10}=1$
$x=\frac{6–4}{10}=0.2$

Thus, the roots are $x=1$ and $x=0.2$.

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Example 14: Solve $x^2–9x+18=0$ by factoring.

Solution:
Find two numbers that multiply to 18 and add to -9. These numbers are -6 and -3.

$x^2–9x+18=(x–6)(x–3)$

Thus, the roots are $x=6$ and $x=3$.

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Example 15: Solve $x^2+4x–5=0$ by factoring.

Solution:
Find two numbers that multiply to -5 and add to 4. These numbers are 5 and -1.

$x^2+4x–5=(x+5)(x–1)$

Thus, the roots are $x=-5$ and $x=1$.

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Example 16: Solve $4x^2+12x+9=0$ by factoring.

Solution:
This is a perfect square trinomial:

$4x^2+12x+9=(2x+3{)}^2$

Thus, the root is $x=-1.5$ (with multiplicity 2).

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Example 17: Solve $3x^2–2x–8=0$ using the quadratic formula.

Step 1: Identify $a=3$, $b=-2$, and $c=-8$.

Step 2: Use the quadratic formula:
$x=\frac{-(-2)\pm \sqrt{(-2{)}^2–4(3)(-8)}}{2(3)}$

$x=\frac{2\pm \sqrt{4+96}}{6}$

$x=\frac{2\pm \sqrt{100}}{6}$

$x=\frac{2\pm 10}{6}$

Step 3: Solve for $x$:
$x=\frac{2+10}{6}=2$
$x=\frac{2–10}{6}=-\frac{4}{3}$

Thus, the roots are $x=2$ and $x=-\frac{4}{3}$.

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Example 18: Solve $x^2–16x+64=0$ by factoring.

Solution:
This is a perfect square trinomial:

$x^2–16x+64=(x–8{)}^2$

Thus, the root is $x=8$ (with multiplicity 2).

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Example 19: Solve $2x^2–4x–6=0$ by factoring.

Solution:
Factor out the GCF:

$2(x^2–2x–3)=0$

Now factor the trinomial:

$2(x–3)(x+1)=0$

Thus, the roots are $x=3$ and $x=-1$.

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Example 20: Solve $x^2+5x+6=0$ by factoring.

Solution:
Find two numbers that multiply to 6 and add to 5. These numbers are 3 and 2.

$x^2+5x+6=(x+3)(x+2)$

Thus, the roots are $x=-3$ and $x=-2$.

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Example 21: Solve $x^2–7x+12=0$ by factoring.

Solution:
Find two numbers that multiply to 12 and add to -7. These numbers are -3 and -4.

$x^2–7x+12=(x–3)(x–4)$

Thus, the roots are $x=3$ and $x=4$.

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Example 22: Solve $2x^2+3x–5=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=3$, and $c=-5$.

Step 2: Use the quadratic formula:
$x=\frac{-3\pm \sqrt{3^2–4(2)(-5)}}{2(2)}$

$x=\frac{-3\pm \sqrt{9+40}}{4}$

$x=\frac{-3\pm \sqrt{49}}{4}$

$x=\frac{-3\pm 7}{4}$

Step 3: Solve for $x$:
$x=\frac{-3+7}{4}=1$
$x=\frac{-3–7}{4}=-\frac{5}{2}$

Thus, the roots are $x=1$ and $x=-\frac{5}{2}$.

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Example 23: Solve $x^2–9x+20=0$ by factoring.

Solution:
Find two numbers that multiply to 20 and add to -9. These numbers are -5 and -4.

$x^2–9x+20=(x–5)(x–4)$

Thus, the roots are $x=5$ and $x=4$.

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Example 24: Solve $x^2+12x+36=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:

$x^2+12x+36=(x+6{)}^2$

Thus, the root is $x=-6$ (with multiplicity 2).

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Example 25: Solve $3x^2–5x–2=0$ using the quadratic formula.

Step 1: Identify $a=3$, $b=-5$, and $c=-2$.

Step 2: Use the quadratic formula:
$x=\frac{-(-5)\pm \sqrt{(-5{)}^2–4(3)(-2)}}{2(3)}$

$x=\frac{5\pm \sqrt{25+24}}{6}$

$x=\frac{5\pm \sqrt{49}}{6}$

$x=\frac{5\pm 7}{6}$

Step 3: Solve for $x$:
$x=\frac{5+7}{6}=2$
$x=\frac{5–7}{6}=-\frac{1}{3}$

Thus, the roots are $x=2$ and $x=-\frac{1}{3}$.

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Example 26: Solve $x^2–10x+24=0$ by factoring.

Solution:
Find two numbers that multiply to 24 and add to -10. These numbers are -6 and -4.

$x^2–10x+24=(x–6)(x–4)$

Thus, the roots are $x=6$ and $x=4$.

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Example 27: Solve $x^2–49=0$ by factoring.

Solution:
Recognize this as a difference of squares:

$x^2–49=(x+7)(x–7)$

Thus, the roots are $x=7$ and $x=-7$.

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Example 28: Solve $4x^2–25=0$ by factoring.

Solution:
Recognize this as a difference of squares:

$4x^2–25=(2x+5)(2x–5)$

Thus, the roots are $x=\frac{5}{2}$ and $x=-\frac{5}{2}$.

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Example 29: Solve $x^2–16=0$ by factoring.

Solution:
Recognize this as a difference of squares:

$x^2–16=(x+4)(x–4)$

Thus, the roots are $x=4$ and $x=-4$.

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Example 30: Solve $5x^2–30x+45=0$ by factoring.

Solution:
Factor out the GCF:
$5(x^2–6x+9)=0$

Now factor the trinomial:
$5(x–3{)}^2=0$

Thus, the root is $x=3$ (with multiplicity 2).

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Example 31: Solve $2x^2+5x–3=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=5$, and $c=-3$.

Step 2: Use the quadratic formula:
$x=\frac{-5\pm \sqrt{5^2–4(2)(-3)}}{2(2)}$

$x=\frac{-5\pm \sqrt{25+24}}{4}$

$x=\frac{-5\pm \sqrt{49}}{4}$

$x=\frac{-5\pm 7}{4}$

Step 3: Solve for $x$:
$x=\frac{-5+7}{4}=\frac{2}{4}=\frac{1}{2}$
$x=\frac{-5–7}{4}=\frac{-12}{4}=-3$

Thus, the roots are $x=\frac{1}{2}$ and $x=-3$.

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Example 32: Solve $x^2–8x+15=0$ by factoring.

Solution:
Find two numbers that multiply to 15 and add to -8. These numbers are -5 and -3.

$x^2–8x+15=(x–5)(x–3)$

Thus, the roots are $x=5$ and $x=3$.

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Example 33: Solve $x^2+2x–15=0$ by factoring.

Solution:
Find two numbers that multiply to -15 and add to 2. These numbers are 5 and -3.

$x^2+2x–15=(x+5)(x–3)$

Thus, the roots are $x=-5$ and $x=3$.

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Example 34: Solve $3x^2–12x+9=0$ by factoring.

Solution:
Factor out the GCF:
$3(x^2–4x+3)=0$

Now factor the trinomial:
$3(x–3)(x–1)=0$

Thus, the roots are $x=3$ and $x=1$.

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Example 35: Solve $x^2+6x–16=0$ by factoring.

Solution:
Find two numbers that multiply to -16 and add to 6. These numbers are 8 and -2.

$x^2+6x–16=(x+8)(x–2)$

Thus, the roots are $x=-8$ and $x=2$.

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Example 36: Solve $4x^2–9=0$ by factoring.

Solution:
Recognize this as a difference of squares:

$4x^2–9=(2x+3)(2x–3)$

Thus, the roots are $x=\frac{3}{2}$ and $x=-\frac{3}{2}$.

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Example 37: Solve $x^2+10x+21=0$ by factoring.

Solution:
Find two numbers that multiply to 21 and add to 10. These numbers are 7 and 3.

$x^2+10x+21=(x+7)(x+3)$

Thus, the roots are $x=-7$ and $x=-3$.

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Example 38: Solve $x^2–10x+24=0$ by factoring.

Solution:
Find two numbers that multiply to 24 and add to -10. These numbers are -6 and -4.

$x^2–10x+24=(x–6)(x–4)$

Thus, the roots are $x=6$ and $x=4$.

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Example 39: Solve $6x^2–5x–4=0$ using the quadratic formula.

Step 1: Identify $a=6$, $b=-5$, and $c=-4$.

Step 2: Use the quadratic formula:
$x=\frac{-(-5)\pm \sqrt{(-5{)}^2–4(6)(-4)}}{2(6)}$

$x=\frac{5\pm \sqrt{25+96}}{12}$

$x=\frac{5\pm \sqrt{121}}{12}$

$x=\frac{5\pm 11}{12}$

Step 3: Solve for $x$:
$x=\frac{5+11}{12}=\frac{16}{12}=\frac{4}{3}$
$x=\frac{5–11}{12}=\frac{-6}{12}=-\frac{1}{2}$

Thus, the roots are $x=\frac{4}{3}$ and $x=-\frac{1}{2}$.

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Example 40: Solve $x^2–5x+6=0$ by factoring.

Solution:
Find two numbers that multiply to 6 and add to -5. These numbers are -2 and -3.

$x^2–5x+6=(x–2)(x–3)$

Thus, the roots are $x=2$ and $x=3$.

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Example 41: Solve $x^2+8x+15=0$ by factoring.

Solution:
Find two numbers that multiply to 15 and add to 8. These numbers are 5 and 3.

$x^2+8x+15=(x+5)(x+3)$

Thus, the roots are $x=-5$ and $x=-3$.

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Example 42: Solve $2x^2–4x–6=0$ by factoring.

Solution:
Factor out the GCF:
$2(x^2–2x–3)=0$

Now factor the quadratic:
$2(x–3)(x+1)=0$

Thus, the roots are $x=3$ and $x=-1$.

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Example 43: Solve $x^2–4x+4=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:
$x^2–4x+4=(x–2{)}^2$

Thus, the root is $x=2$ (with multiplicity 2).

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Example 44: Solve $x^2+6x+8=0$ by factoring.

Solution:
Find two numbers that multiply to 8 and add to 6. These numbers are 4 and 2.

$x^2+6x+8=(x+4)(x+2)$

Thus, the roots are $x=-4$ and $x=-2$.

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Example 45: Solve $3x^2–12x+9=0$ by factoring.

Solution:
Factor out the GCF:
$3(x^2–4x+3)=0$

Now factor the quadratic:
$3(x–3)(x–1)=0$

Thus, the roots are $x=3$ and $x=1$.

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Example 46: Solve $5x^2+14x–3=0$ using the quadratic formula.

Step 1: Identify $a=5$, $b=14$, and $c=-3$.

Step 2: Use the quadratic formula:
$x=\frac{-14\pm \sqrt{{14}^2–4(5)(-3)}}{2(5)}$

$x=\frac{-14\pm \sqrt{196+60}}{10}$

$x=\frac{-14\pm \sqrt{256}}{10}$

$x=\frac{-14\pm 16}{10}$

Step 3: Solve for $x$:
$x=\frac{-14+16}{10}=\frac{2}{10}=\frac{1}{5}$
$x=\frac{-14–16}{10}=\frac{-30}{10}=-3$

Thus, the roots are $x=\frac{1}{5}$ and $x=-3$.

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Example 47: Solve $x^2–10x+21=0$ by factoring.

Solution:
Find two numbers that multiply to 21 and add to -10. These numbers are -7 and -3.

$x^2–10x+21=(x–7)(x–3)$

Thus, the roots are $x=7$ and $x=3$.

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Example 48: Solve $x^2+3x–18=0$ by factoring.

Solution:
Find two numbers that multiply to -18 and add to 3. These numbers are 6 and -3.

$x^2+3x–18=(x+6)(x–3)$

Thus, the roots are $x=-6$ and $x=3$.

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Example 49: Solve $2x^2+9x+4=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=9$, and $c=4$.

Step 2: Use the quadratic formula:
$x=\frac{-9\pm \sqrt{9^2–4(2)(4)}}{2(2)}$

$x=\frac{-9\pm \sqrt{81–32}}{4}$

$x=\frac{-9\pm \sqrt{49}}{4}$

$x=\frac{-9\pm 7}{4}$

Step 3: Solve for $x$:
$x=\frac{-9+7}{4}=\frac{-2}{4}=-\frac{1}{2}$
$x=\frac{-9–7}{4}=\frac{-16}{4}=-4$

Thus, the roots are $x=-\frac{1}{2}$ and $x=-4$.

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Example 50: Solve $x^2–2x–15=0$ by factoring.

Solution:
Find two numbers that multiply to -15 and add to -2. These numbers are -5 and 3.

$x^2–2x–15=(x–5)(x+3)$

Thus, the roots are $x=5$ and $x=-3$.

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Example 51: Solve $x^2+10x+24=0$ by factoring.

Solution:
Find two numbers that multiply to 24 and add to 10. These numbers are 6 and 4.

$x^2+10x+24=(x+6)(x+4)$

Thus, the roots are $x=-6$ and $x=-4$.

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Example 52: Solve $2x^2–7x+3=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=-7$, and $c=3$.

Step 2: Use the quadratic formula:
$x=\frac{-(-7)\pm \sqrt{(-7{)}^2–4(2)(3)}}{2(2)}$

$x=\frac{7\pm \sqrt{49–24}}{4}$

$x=\frac{7\pm \sqrt{25}}{4}$

$x=\frac{7\pm 5}{4}$

Step 3: Solve for $x$:
$x=\frac{7+5}{4}=3$
$x=\frac{7–5}{4}=\frac{1}{2}$

Thus, the roots are $x=3$ and $x=\frac{1}{2}$.

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Example 53: Solve $x^2+9x+20=0$ by factoring.

Solution:
Find two numbers that multiply to 20 and add to 9. These numbers are 4 and 5.

$x^2+9x+20=(x+4)(x+5)$

Thus, the roots are $x=-4$ and $x=-5$.

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Example 54: Solve $4x^2–8x+4=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:
$4x^2–8x+4=(2x–2{)}^2$

Thus, the root is $x=1$ (with multiplicity 2).

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Example 55: Solve $x^2–5x+6=0$ by factoring.

Solution:
Find two numbers that multiply to 6 and add to -5. These numbers are -3 and -2.

$x^2–5x+6=(x–3)(x–2)$

Thus, the roots are $x=3$ and $x=2$.

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Example 57: Solve $x^2–6x+8=0$ by factoring.

Solution:
Find two numbers that multiply to 8 and add to -6. These numbers are -4 and -2.

$x^2–6x+8=(x–4)(x–2)$

Thus, the roots are $x=4$ and $x=2$.

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Example 58: Solve $2x^2+3x–2=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=3$, and $c=-2$.

Step 2: Use the quadratic formula:
$x=\frac{-3\pm \sqrt{3^2–4(2)(-2)}}{2(2)}$

$x=\frac{-3\pm \sqrt{9+16}}{4}$

$x=\frac{-3\pm \sqrt{25}}{4}$

$x=\frac{-3\pm 5}{4}$

Step 3: Solve for $x$:
$x=\frac{-3+5}{4}=\frac{2}{4}=\frac{1}{2}$
$x=\frac{-3–5}{4}=\frac{-8}{4}=-2$

Thus, the roots are $x=\frac{1}{2}$ and $x=-2$.

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Example 59: Solve $x^2+7x+12=0$ by factoring.

Solution:
Find two numbers that multiply to 12 and add to 7. These numbers are 4 and 3.

$x^2+7x+12=(x+4)(x+3)$

Thus, the roots are $x=-4$ and $x=-3$.

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Example 60: Solve $x^2–8x+16=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:
$x^2–8x+16=(x–4{)}^2$

Thus, the root is $x=4$ (with multiplicity 2).

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Example 61: Solve $x^2–3x–10=0$ by factoring.

Solution:
Find two numbers that multiply to -10 and add to -3. These numbers are -5 and 2.

$x^2–3x–10=(x–5)(x+2)$

Thus, the roots are $x=5$ and $x=-2$.

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Example 62: Solve $5x^2–4x–1=0$ using the quadratic formula.

Step 1: Identify $a=5$, $b=-4$, and $c=-1$.

Step 2: Use the quadratic formula:
$x=\frac{-(-4)\pm \sqrt{(-4{)}^2–4(5)(-1)}}{2(5)}$

$x=\frac{4\pm \sqrt{16+20}}{10}$

$x=\frac{4\pm \sqrt{36}}{10}$

$x=\frac{4\pm 6}{10}$

Step 3: Solve for $x$:
$x=\frac{4+6}{10}=1$
$x=\frac{4–6}{10}=-\frac{1}{5}$

Thus, the roots are $x=1$ and $x=-\frac{1}{5}$.

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Example 63: Solve $x^2–4x–5=0$ by factoring.

Solution:
Find two numbers that multiply to -5 and add to -4. These numbers are -5 and 1.

$x^2–4x–5=(x–5)(x+1)$

Thus, the roots are $x=5$ and $x=-1$.

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Example 64: Solve $6x^2–5x+1=0$ using the quadratic formula.

Step 1: Identify $a=6$, $b=-5$, and $c=1$.

Step 2: Use the quadratic formula:
$x=\frac{-(-5)\pm \sqrt{(-5{)}^2–4(6)(1)}}{2(6)}$

$x=\frac{5\pm \sqrt{25–24}}{12}$

$x=\frac{5\pm \sqrt{1}}{12}$

$x=\frac{5\pm 1}{12}$

Step 3: Solve for $x$:
$x=\frac{5+1}{12}=\frac{6}{12}=\frac{1}{2}$
$x=\frac{5–1}{12}=\frac{4}{12}=\frac{1}{3}$

Thus, the roots are $x=\frac{1}{2}$ and $x=\frac{1}{3}$.

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Example 65: Solve $x^2+10x+16=0$ by factoring.

Solution:
Find two numbers that multiply to 16 and add to 10. These numbers are 8 and 2.

$x^2+10x+16=(x+8)(x+2)$

Thus, the roots are $x=-8$ and $x=-2$.

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Example 66: Solve $2x^2–11x+12=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=-11$, and $c=12$.

Step 2: Use the quadratic formula:
$x=\frac{-(-11)\pm \sqrt{(-11{)}^2–4(2)(12)}}{2(2)}$

$x=\frac{11\pm \sqrt{121–96}}{4}$

$x=\frac{11\pm \sqrt{25}}{4}$

$x=\frac{11\pm 5}{4}$

Step 3: Solve for $x$:
$x=\frac{11+5}{4}=4$
$x=\frac{11–5}{4}=\frac{6}{4}=\frac{3}{2}$

Thus, the roots are $x=4$ and $x=\frac{3}{2}$.

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Example 67: Solve $x^2–6x+9=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:
$x^2–6x+9=(x–3{)}^2$

Thus, the root is $x=3$ (with multiplicity 2).

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Example 68: Solve $x^2–9x+14=0$ by factoring.

Solution:
Find two numbers that multiply to 14 and add to -9. These numbers are -7 and -2.

$x^2–9x+14=(x–7)(x–2)$

Thus, the roots are $x=7$ and $x=2$.

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Example 69: Solve $4x^2–16x+15=0$ using the quadratic formula.

Step 1: Identify $a=4$, $b=-16$, and $c=15$.

Step 2: Use the quadratic formula:
$x=\frac{-(-16)\pm \sqrt{(-16{)}^2–4(4)(15)}}{2(4)}$

$x=\frac{16\pm \sqrt{256–240}}{8}$

$x=\frac{16\pm \sqrt{16}}{8}$

$x=\frac{16\pm 4}{8}$

Step 3: Solve for $x$:
$x=\frac{16+4}{8}=\frac{20}{8}=2.5$
$x=\frac{16–4}{8}=\frac{12}{8}=1.5$

Thus, the roots are $x=2.5$ and $x=1.5$.

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Example 70: Solve $x^2–8x+15=0$ by factoring.

Solution:
Find two numbers that multiply to 15 and add to -8. These numbers are -5 and -3.

$x^2–8x+15=(x–5)(x–3)$

Thus, the roots are $x=5$ and $x=3$.

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Example 71: Solve $x^2+2x–8=0$ by factoring.

Solution:
Find two numbers that multiply to -8 and add to 2. These numbers are 4 and -2.

$x^2+2x–8=(x+4)(x–2)$

Thus, the roots are $x=-4$ and $x=2$.

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Example 72: Solve $2x^2–3x–2=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=-3$, and $c=-2$.

Step 2: Use the quadratic formula:
$x=\frac{-(-3)\pm \sqrt{(-3{)}^2–4(2)(-2)}}{2(2)}$

$x=\frac{3\pm \sqrt{9+16}}{4}$

$x=\frac{3\pm \sqrt{25}}{4}$

$x=\frac{3\pm 5}{4}$

Step 3: Solve for $x$:
$x=\frac{3+5}{4}=2$
$x=\frac{3–5}{4}=-\frac{1}{2}$

Thus, the roots are $x=2$ and $x=-\frac{1}{2}$.

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Example 73: Solve $x^2–7x+12=0$ by factoring.

Solution:
Find two numbers that multiply to 12 and add to -7. These numbers are -3 and -4.

$x^2–7x+12=(x–3)(x–4)$

Thus, the roots are $x=3$ and $x=4$.

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Example 74: Solve $x^2+5x+6=0$ by factoring.

Solution:
Find two numbers that multiply to 6 and add to 5. These numbers are 3 and 2.

$x^2+5x+6=(x+3)(x+2)$

Thus, the roots are $x=-3$ and $x=-2$.

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Example 75: Solve $3x^2+8x–16=0$ using the quadratic formula.

Step 1: Identify $a=3$, $b=8$, and $c=-16$.

Step 2: Use the quadratic formula:
$x=\frac{-8\pm \sqrt{8^2–4(3)(-16)}}{2(3)}$

$x=\frac{-8\pm \sqrt{64+192}}{6}$

$x=\frac{-8\pm \sqrt{256}}{6}$

$x=\frac{-8\pm 16}{6}$

Step 3: Solve for $x$:
$x=\frac{-8+16}{6}=\frac{8}{6}=\frac{4}{3}$
$x=\frac{-8–16}{6}=\frac{-24}{6}=-4$

Thus, the roots are $x=\frac{4}{3}$ and $x=-4$.

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Example 76: Solve $x^2–4x–21=0$ by factoring.

Solution:
Find two numbers that multiply to -21 and add to -4. These numbers are -7 and 3.

$x^2–4x–21=(x–7)(x+3)$

Thus, the roots are $x=7$ and $x=-3$.

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Example 77: Solve $x^2+4x–12=0$ by factoring.

Solution:
Find two numbers that multiply to -12 and add to 4. These numbers are 6 and -2.

$x^2+4x–12=(x+6)(x–2)$

Thus, the roots are $x=-6$ and $x=2$.

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Example 78: Solve $4x^2–9=0$ by factoring.

Solution:
Recognize this as a difference of squares:

$4x^2–9=(2x+3)(2x–3)$

Thus, the roots are $x=\frac{3}{2}$ and $x=-\frac{3}{2}$.

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Example 79: Solve $x^2–8x+16=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:
$x^2–8x+16=(x–4{)}^2$

Thus, the root is $x=4$ (with multiplicity 2).

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Example 80: Solve $x^2–11x+28=0$ by factoring.

Solution:
Find two numbers that multiply to 28 and add to -11. These numbers are -7 and -4.

$x^2–11x+28=(x–7)(x–4)$

Thus, the roots are $x=7$ and $x=4$.

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Example 81: Solve $2x^2+3x–2=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=3$, and $c=-2$.

Step 2: Use the quadratic formula:
$x=\frac{-3\pm \sqrt{3^2–4(2)(-2)}}{2(2)}$

$x=\frac{-3\pm \sqrt{9+16}}{4}$

$x=\frac{-3\pm \sqrt{25}}{4}$

$x=\frac{-3\pm 5}{4}$

Step 3: Solve for $x$:
$x=\frac{-3+5}{4}=\frac{2}{4}=\frac{1}{2}$
$x=\frac{-3–5}{4}=\frac{-8}{4}=-2$

Thus, the roots are $x=\frac{1}{2}$ and $x=-2$.

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Example 82: Solve $x^2+5x–14=0$ by factoring.

Solution:
Find two numbers that multiply to -14 and add to 5. These numbers are 7 and -2.

$x^2+5x–14=(x+7)(x–2)$

Thus, the roots are $x=-7$ and $x=2$.

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Example 83: Solve $3x^2–2x–1=0$ using the quadratic formula.

Step 1: Identify $a=3$, $b=-2$, and $c=-1$.

Step 2: Use the quadratic formula:
$x=\frac{-(-2)\pm \sqrt{(-2{)}^2–4(3)(-1)}}{2(3)}$

$x=\frac{2\pm \sqrt{4+12}}{6}$

$x=\frac{2\pm \sqrt{16}}{6}$

$x=\frac{2\pm 4}{6}$

Step 3: Solve for $x$:
$x=\frac{2+4}{6}=1$
$x=\frac{2–4}{6}=-\frac{1}{3}$

Thus, the roots are $x=1$ and $x=-\frac{1}{3}$.

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Example 84: Solve $x^2+4x–21=0$ by factoring.

Solution:
Find two numbers that multiply to -21 and add to 4. These numbers are 7 and -3.

$x^2+4x–21=(x+7)(x–3)$

Thus, the roots are $x=-7$ and $x=3$.

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Example 85: Solve $2x^2–5x–3=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=-5$, and $c=-3$.

Step 2: Use the quadratic formula:
$x=\frac{-(-5)\pm \sqrt{(-5{)}^2–4(2)(-3)}}{2(2)}$

$x=\frac{5\pm \sqrt{25+24}}{4}$

$x=\frac{5\pm \sqrt{49}}{4}$

$x=\frac{5\pm 7}{4}$

Step 3: Solve for $x$:
$x=\frac{5+7}{4}=3$
$x=\frac{5–7}{4}=-\frac{1}{2}$

Thus, the roots are $x=3$ and $x=-\frac{1}{2}$.

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Example 86: Solve $x^2+7x+10=0$ by factoring.

Solution:
Find two numbers that multiply to 10 and add to 7. These numbers are 5 and 2.

$x^2+7x+10=(x+5)(x+2)$

Thus, the roots are $x=-5$ and $x=-2$.

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Example 87: Solve $x^2+7x+10=0$ by factoring.

Solution:
Find two numbers that multiply to 10 and add to 7. These numbers are 5 and 2.

$x^2+7x+10=(x+5)(x+2)$

Thus, the roots are $x=-5$ and $x=-2$.

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Example 88: Solve $x^2–12x+36=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:

$x^2–12x+36=(x–6{)}^2$

Thus, the root is $x=6$ (with multiplicity 2).

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Example 89: Solve $4x^2–25=0$ by factoring.

Solution:
Recognize this as a difference of squares:

$4x^2–25=(2x+5)(2x–5)$

Thus, the roots are $x=\frac{5}{2}$ and $x=-\frac{5}{2}$.

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Example 90: Solve $x^2–9x+20=0$ by factoring.

Solution:
Find two numbers that multiply to 20 and add to -9. These numbers are -5 and -4.

$x^2–9x+20=(x–5)(x–4)$

Thus, the roots are $x=5$ and $x=4$.

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Example 91: Solve $2x^2–x–6=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=-1$, and $c=-6$.

Step 2: Use the quadratic formula:
$x=\frac{-(-1)\pm \sqrt{(-1{)}^2–4(2)(-6)}}{2(2)}$

$x=\frac{1\pm \sqrt{1+48}}{4}$

$x=\frac{1\pm \sqrt{49}}{4}$

$x=\frac{1\pm 7}{4}$

Step 3: Solve for $x$:
$x=\frac{1+7}{4}=\frac{8}{4}=2$
$x=\frac{1–7}{4}=\frac{-6}{4}=-\frac{3}{2}$

Thus, the roots are $x=2$ and $x=-\frac{3}{2}$.

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Example 92: Solve $x^2+2x–15=0$ by factoring.

Solution:
Find two numbers that multiply to -15 and add to 2. These numbers are 5 and -3.

$x^2+2x–15=(x+5)(x–3)$

Thus, the roots are $x=-5$ and $x=3$.

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Example 93: Solve $3x^2+4x–4=0$ using the quadratic formula.

Step 1: Identify $a=3$, $b=4$, and $c=-4$.

Step 2: Use the quadratic formula:
$x=\frac{-4\pm \sqrt{4^2–4(3)(-4)}}{2(3)}$

$x=\frac{-4\pm \sqrt{16+48}}{6}$

$x=\frac{-4\pm \sqrt{64}}{6}$

$x=\frac{-4\pm 8}{6}$

Step 3: Solve for $x$:
$x=\frac{-4+8}{6}=\frac{4}{6}=\frac{2}{3}$
$x=\frac{-4–8}{6}=\frac{-12}{6}=-2$

Thus, the roots are $x=\frac{2}{3}$ and $x=-2$.

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Example 94: Solve $x^2–5x+6=0$ by factoring.

Solution:
Find two numbers that multiply to 6 and add to -5. These numbers are -3 and -2.

$x^2–5x+6=(x–3)(x–2)$

Thus, the roots are $x=3$ and $x=2$.

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Example 95: Solve $5x^2–9x+4=0$ by factoring.

Solution:
Factor the quadratic expression:
$5x^2–9x+4=(5x–4)(x–1)$

Thus, the roots are $x=\frac{4}{5}$ and $x=1$.

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Example 96: Solve $4x^2–12x+9=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:
$4x^2–12x+9=(2x–3{)}^2$

Thus, the root is $x=\frac{3}{2}$ (with multiplicity 2).

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Example 97: Solve $x^2+4x+4=0$ by factoring.

Solution:
Recognize this as a perfect square trinomial:
$x^2+4x+4=(x+2{)}^2$

Thus, the root is $x=-2$ (with multiplicity 2).

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Example 98: Solve $2x^2–7x+3=0$ using the quadratic formula.

Step 1: Identify $a=2$, $b=-7$, and $c=3$.

Step 2: Use the quadratic formula:
$x=\frac{-(-7)\pm \sqrt{(-7{)}^2–4(2)(3)}}{2(2)}$

$x=\frac{7\pm \sqrt{49–24}}{4}$

$x=\frac{7\pm \sqrt{25}}{4}$

$x=\frac{7\pm 5}{4}$

Step 3: Solve for $x$:
$x=\frac{7+5}{4}=3$
$x=\frac{7–5}{4}=\frac{1}{2}$

Thus, the roots are $x=3$ and $x=\frac{1}{2}$.

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Example 99: Solve $x^2–2x–3=0$ by factoring.

Solution:
Find two numbers that multiply to -3 and add to -2. These numbers are -3 and 1.

$x^2–2x–3=(x–3)(x+1)$

Thus, the roots are $x=3$ and $x=-1$.

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Example 100: Solve $6x^2–5x–4=0$ using the quadratic formula.

Step 1: Identify $a=6$, $b=-5$, and $c=-4$.

Step 2: Use the quadratic formula:
$x=\frac{-(-5)\pm \sqrt{(-5{)}^2–4(6)(-4)}}{2(6)}$

$x=\frac{5\pm \sqrt{25+96}}{12}$

$x=\frac{5\pm \sqrt{121}}{12}$

$x=\frac{5\pm 11}{12}$

Step 3: Solve for $x$:
$x=\frac{5+11}{12}=\frac{16}{12}=\frac{4}{3}$
$x=\frac{5–11}{12}=\frac{-6}{12}=-\frac{1}{2}$

Thus, the roots are $x=\frac{4}{3}$ and $x=-\frac{1}{2}$.