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

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

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 $a = - 1$ , $b = 0$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (- 1 \cdot 16)}}{2 \cdot - 1}$

Step 3

Simplify.

Simplify the numerator.

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot - 1 \cdot 16}}{2 \cdot - 1}$

Multiply $- 4 \cdot - 1 \cdot 16$ .

Multiply $- 4$ by $- 1$ .

$x = \frac{0 \pm \sqrt{0 + 4 \cdot 16}}{2 \cdot - 1}$

Multiply $4$ by $16$ .

$x = \frac{0 \pm \sqrt{0 + 64}}{2 \cdot - 1}$

Add $0$ and $64$ .

$x = \frac{0 \pm \sqrt{64}}{2 \cdot - 1}$

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

$x = \frac{0 \pm \sqrt{8^{2}}}{2 \cdot - 1}$

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

$x = \frac{0 \pm 8}{2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$x = \frac{0 \pm 8}{- 2}$

Simplify $\frac{0 \pm 8}{- 2}$ .

$x = \pm 4$

Step 4

The final answer is the combination of both solutions.

$x = 4, - 4$