Solve Using the Quadratic Formula y^2+16y+64=0

$y^{2} + 16 y + 64 = 0$

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 $a = 1$ , $b = 16$ , and $c = 64$ into the quadratic formula and solve for $y$ .

$\frac{- 16 \pm \sqrt{16^{2} - 4 \cdot (1 \cdot 64)}}{2 \cdot 1}$

Step 3

Simplify.

Simplify the numerator.

Raise $16$ to the power of $2$ .

$y = \frac{- 16 \pm \sqrt{256 - 4 \cdot 1 \cdot 64}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 64$ .

Multiply $- 4$ by $1$ .

$y = \frac{- 16 \pm \sqrt{256 - 4 \cdot 64}}{2 \cdot 1}$

Multiply $- 4$ by $64$ .

$y = \frac{- 16 \pm \sqrt{256 - 256}}{2 \cdot 1}$

Subtract $256$ from $256$ .

$y = \frac{- 16 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$y = \frac{- 16 \pm \sqrt{0^{2}}}{2 \cdot 1}$

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

$y = \frac{- 16 \pm 0}{2 \cdot 1}$

$- 16$ plus or minus $0$ is $- 16$ .

$y = \frac{- 16}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{- 16}{2}$

Divide $- 16$ by $2$ .

$y = - 8$

Step 4

The final answer is the combination of both solutions.

$y = - 8$ Double roots