Solve Using the Quadratic Formula y^2-8y+32=0

$y^{2} - 8 y + 32 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 8$ , and $c = 32$ into the quadratic formula and solve for $y$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 32)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 8$ to the power of $2$ .

$y = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 32}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 32$ .

Multiply $- 4$ by $1$ .

$y = \frac{8 \pm \sqrt{64 - 4 \cdot 32}}{2 \cdot 1}$

Multiply $- 4$ by $32$ .

$y = \frac{8 \pm \sqrt{64 - 128}}{2 \cdot 1}$

Subtract $128$ from $64$ .

$y = \frac{8 \pm \sqrt{- 64}}{2 \cdot 1}$

Rewrite $- 64$ as $- 1 (64)$ .

$y = \frac{8 \pm \sqrt{- 1 \cdot 64}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (64)}$ as $\sqrt{- 1} \cdot \sqrt{64}$ .

$y = \frac{8 \pm \sqrt{- 1} \cdot \sqrt{64}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{8 \pm i \cdot \sqrt{64}}{2 \cdot 1}$

Rewrite $64$ as $8^{2}$ .

$y = \frac{8 \pm i \cdot \sqrt{8^{2}}}{2 \cdot 1}$

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

$y = \frac{8 \pm i \cdot 8}{2 \cdot 1}$

Move $8$ to the left of $i$ .

$y = \frac{8 \pm 8 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{8 \pm 8 i}{2}$

Simplify $\frac{8 \pm 8 i}{2}$ .

$y = 4 \pm 4 i$