Find the Roots (Zeros) m(x)=-x^2-27

$m (x) = - x^{2} - 27$

Set $- x^{2} - 27$ equal to $0$ .

$- x^{2} - 27 = 0$

Solve for $x$ .

Add $27$ to both sides of the equation.

$- x^{2} = 27$

Divide each term in $- x^{2} = 27$ by $- 1$ and simplify.

Divide each term in $- x^{2} = 27$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{27}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{27}{- 1}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{27}{- 1}$

Simplify the right side.

Divide $27$ by $- 1$ .

$x^{2} = - 27$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 27}$

Simplify $\pm \sqrt{- 27}$ .

Rewrite $- 27$ as $- 1 (27)$ .

$x = \pm \sqrt{- 1 (27)}$

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{27}$

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

$x = \pm i \cdot \sqrt{27}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$x = \pm i \cdot \sqrt{9 (3)}$

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2} \cdot 3}$

Pull terms out from under the radical.

$x = \pm i \cdot (3 \sqrt{3})$

Move $3$ to the left of $i$ .

$x = \pm 3 i \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 3 i \sqrt{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3 i \sqrt{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3 i \sqrt{3}, - 3 i \sqrt{3}$