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

$- x^{2} + 16 x - 64 = 0$

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 $x$ .

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

Step 3

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $- 1$ .

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

Multiply $4$ by $- 64$ .

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

Subtract $256$ from $256$ .

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

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

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

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

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

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

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

Multiply $2$ by $- 1$ .

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

Divide $- 16$ by $- 2$ .

$x = 8$

Step 4

The final answer is the combination of both solutions.

$x = 8$ Double roots