# How do you solve x^2+6x+13=0 by completing the square?

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#### Explanation:

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Oct 31, 2016

$x = - 3 \pm 2 i$

#### Explanation:

Solve by completing the square.

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

Move the constant to the right side by subtracting 13 from each side.

${x}^{2} + 6 x \textcolor{w h i t e}{a a a a} = - 13$

Divide the coefficient of the $x$ term by $2$.

${x}^{2} + \textcolor{red}{6} x \textcolor{w h i t e}{a a a a} = - 13$

$\frac{\textcolor{red}{6}}{2} = \textcolor{b l u e}{3}$

Square the result and add it to both sides.

${\textcolor{b l u e}{3}}^{2} = \textcolor{m a \ge n t a}{9}$

${x}^{2} + 6 x + \textcolor{m a \ge n t a}{9} = - 13 + \textcolor{m a \ge n t a}{9}$

Factor the left side and simplify the right side.

$\left(x + \textcolor{b l u e}{3}\right) \left(x + \textcolor{b l u e}{3}\right) = - 4$

Rewrite as the square of the binomial. Note that the $\textcolor{b l u e}{3}$ in the binomial is the same value $\textcolor{b l u e}{3}$ that resulted from dividing the coefficient of the $x$ term by $2$.

${\left(x + \textcolor{b l u e}{3}\right)}^{2} = - 4$

Square root both sides and solve for $x$.

$\sqrt{{\left(x + \textcolor{b l u e}{3}\right)}^{2}} = \sqrt{- 4}$

$x + \textcolor{b l u e}{3} = \pm 2 i$

$x = - 3 \pm 2 i$

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