Solve Using the Quadratic Formula (x-2)(x+3)=14

$(x - 2) (x + 3) = 14$

Step 1

Move all terms to the left side of the equation and simplify.

Simplify the left side.

Simplify $(x - 2) (x + 3)$ .

Expand $(x - 2) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$x (x + 3) - 2 (x + 3) = 14$

Apply the distributive property.

$x \cdot x + x \cdot 3 - 2 (x + 3) = 14$

Apply the distributive property.

$x \cdot x + x \cdot 3 - 2 x - 2 \cdot 3 = 14$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$x^{2} + x \cdot 3 - 2 x - 2 \cdot 3 = 14$

Move $3$ to the left of $x$ .

$x^{2} + 3 \cdot x - 2 x - 2 \cdot 3 = 14$

Multiply $- 2$ by $3$ .

$x^{2} + 3 x - 2 x - 6 = 14$

Subtract $2 x$ from $3 x$ .

$x^{2} + x - 6 = 14$

Subtract $14$ from both sides of the equation.

$x^{2} + x - 6 - 14 = 0$

Subtract $14$ from $- 6$ .

$x^{2} + x - 20 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 1$ , and $c = - 20$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot - 20)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 1 \cdot - 20}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 20$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot - 20}}{2 \cdot 1}$

Multiply $- 4$ by $- 20$ .

$x = \frac{- 1 \pm \sqrt{1 + 80}}{2 \cdot 1}$

Add $1$ and $80$ .

$x = \frac{- 1 \pm \sqrt{81}}{2 \cdot 1}$

Rewrite $81$ as $9^{2}$ .

$x = \frac{- 1 \pm \sqrt{9^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 1 \pm 9}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm 9}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 4, - 5$

Solve Using the Quadratic Formula (x-2)(x+3)=14

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