Solve Using the Square Root Property 2x^2+27=-13

$2 x^{2} + 27 = - 13$

Step 1

Move all terms not containing $x$ to the right side of the equation.

Subtract $27$ from both sides of the equation.

$2 x^{2} = - 13 - 27$

Subtract $27$ from $- 13$ .

$2 x^{2} = - 40$

Step 2

Divide each term in $2 x^{2} = - 40$ by $2$ and simplify.

Divide each term in $2 x^{2} = - 40$ by $2$ .

$\frac{2 x^{2}}{2} = \frac{- 40}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x^{2}}{2} = \frac{- 40}{2}$

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

$x^{2} = \frac{- 40}{2}$

Simplify the right side.

Divide $- 40$ by $2$ .

$x^{2} = - 20$

Step 3

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

$x = \pm \sqrt{- 20}$

Step 4

Simplify $\pm \sqrt{- 20}$ .

Rewrite $- 20$ as $- 1 (20)$ .

$x = \pm \sqrt{- 1 (20)}$

Rewrite $\sqrt{- 1 (20)}$ as $\sqrt{- 1} \cdot \sqrt{20}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{20}$

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

$x = \pm i \cdot \sqrt{20}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$x = \pm i \cdot \sqrt{4 (5)}$

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

$x = \pm i \cdot \sqrt{2^{2} \cdot 5}$

Pull terms out from under the radical.

$x = \pm i \cdot (2 \sqrt{5})$

Move $2$ to the left of $i$ .

$x = \pm 2 i \sqrt{5}$

Step 5

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.

$x = 2 i \sqrt{5}$

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

$x = - 2 i \sqrt{5}$

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

$x = 2 i \sqrt{5}, - 2 i \sqrt{5}$

Solve Using the Square Root Property 2x^2+27=-13

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