Find the Roots (Zeros) x^2+16=-12x

$x^{2} + 16 = - 12 x$

Step 1

Add $12 x$ to both sides of the equation.

$x^{2} + 16 + 12 x = 0$

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 $a = 1$ , $b = 12$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

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

Multiply $- 4$ by $1$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot 16}}{2 \cdot 1}$

Multiply $- 4$ by $16$ .

$x = \frac{- 12 \pm \sqrt{144 - 64}}{2 \cdot 1}$

Subtract $64$ from $144$ .

$x = \frac{- 12 \pm \sqrt{80}}{2 \cdot 1}$

Rewrite $80$ as $4^{2} \cdot 5$ .

Factor $16$ out of $80$ .

$x = \frac{- 12 \pm \sqrt{16 (5)}}{2 \cdot 1}$

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

$x = \frac{- 12 \pm \sqrt{4^{2} \cdot 5}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 12 \pm 4 \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 12 \pm 4 \sqrt{5}}{2}$

Simplify $\frac{- 12 \pm 4 \sqrt{5}}{2}$ .

$x = - 6 \pm 2 \sqrt{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - 6 \pm 2 \sqrt{5}$

Decimal Form:

$x = - 1.52786404 \dots, - 10.47213595 \dots$

Step 6