Solve Using the Square Root Property 2x^2+96=0

$2 x^{2} + 96 = 0$

Subtract $96$ from both sides of the equation.

$2 x^{2} = - 96$

Divide each term in $2 x^{2} = - 96$ by $2$ and simplify.

Divide each term in $2 x^{2} = - 96$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{- 96}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{- 96}{2}$

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

$x^{2} = \frac{- 96}{2}$

Simplify the right side.

Divide $- 96$ by $2$ .

$x^{2} = - 48$

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

$x = \pm \sqrt{- 48}$

Simplify $\pm \sqrt{- 48}$ .

Rewrite $- 48$ as $- 1 (48)$ .

$x = \pm \sqrt{- 1 (48)}$

Rewrite $\sqrt{- 1 (48)}$ as $\sqrt{- 1} \cdot \sqrt{48}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{48}$

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

$x = \pm i \cdot \sqrt{48}$

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$x = \pm i \cdot \sqrt{16 (3)}$

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

$x = \pm i \cdot \sqrt{4^{2} \cdot 3}$

Pull terms out from under the radical.

$x = \pm i \cdot (4 \sqrt{3})$

Move $4$ to the left of $i$ .

$x = \pm 4 i \sqrt{3}$

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.

$x = 4 i \sqrt{3}$

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

$x = - 4 i \sqrt{3}$

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

$x = 4 i \sqrt{3}, - 4 i \sqrt{3}$