Solve Using the Quadratic Formula (x+5)^2-12(x+5)+32=0

${(x + 5)}^{2} - 12 (x + 5) + 32 = 0$

Step 1

Simplify the left side.

Simplify ${(x + 5)}^{2} - 12 (x + 5) + 32$ .

Simplify each term.

Apply the distributive property.

${(x + 5)}^{2} - 12 x - 12 \cdot 5 + 32 = 0$

Multiply $- 12$ by $5$ .

${(x + 5)}^{2} - 12 x - 60 + 32 = 0$

Add $- 60$ and $32$ .

${(x + 5)}^{2} - 12 x - 28 = 0$

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 $a = 1$ , $b = - 2$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot - 3)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 3$ .

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 3}}{2 \cdot 1}$

Multiply $- 4$ by $- 3$ .

$x = \frac{2 \pm \sqrt{4 + 12}}{2 \cdot 1}$

Add $4$ and $12$ .

$x = \frac{2 \pm \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

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

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

Multiply $2$ by $1$ .

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

Simplify $\frac{2 \pm 4}{2}$ .

$x = 1 \pm 2$

Step 5

The final answer is the combination of both solutions.

$x = 3, - 1$