Solve by Completing the Square 0=x^2+6x+13

$0 = x^{2} + 6 x + 13$

Simplify the equation into a proper form to complete the square.

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + 6 x + 13 = 0$

Subtract $13$ from both sides of the equation.

$x^{2} + 6 x = - 13$

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(3)}^{2}$

Add the term to each side of the equation.

$x^{2} + 6 x + {(3)}^{2} = - 13 + {(3)}^{2}$

Simplify the equation.

Simplify the left side.

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = - 13 + {(3)}^{2}$

Simplify the right side.

Simplify $- 13 + {(3)}^{2}$ .

Raise $3$ to the power of $2$ .

$x^{2} + 6 x + 9 = - 13 + 9$

Add $- 13$ and $9$ .

$x^{2} + 6 x + 9 = - 4$

Factor the perfect trinomial square into ${(x + 3)}^{2}$ .

${(x + 3)}^{2} = - 4$

Solve the equation for $x$ .

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

$x + 3 = \pm \sqrt{- 4}$

Simplify $\pm \sqrt{- 4}$ .

Rewrite $- 4$ as $- 1 (4)$ .

$x + 3 = \pm \sqrt{- 1 (4)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x + 3 = \pm \sqrt{- 1} \cdot \sqrt{4}$

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

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

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

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

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

$x + 3 = \pm i \cdot 2$

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

$x + 3 = \pm 2 i$

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

Subtract $3$ from both sides of the equation.

$x = \pm 2 i - 3$

Reorder $\pm 2 i$ and $- 3$ .

$x = - 3 \pm 2 i$