# How do you solve the quadratic equation by completing the square: x^2+6x=7?

Aug 1, 2015

${x}_{1 , 2} = - 3 \pm 4$

#### Explanation:

In order to solve this quadratic equation by completing the square, you need to write the left side of the equation as the square of a binomial.

To do that, you need to add a term to both sides of the equaion. More specifically, you need to divide the coefficient of $x$-term by 2 and square the result.

$\frac{6}{2} = 3$, then ${3}^{2} = 9$

Add $9$ to both sides of the equation to get

${x}^{2} + 6 x + 9 = 7 + 9$

The left side of the equation can now be written as

${x}^{2} + 6 x + 9 = {x}^{2} + 2 \cdot \left(3\right) \cdot x + \left({3}^{2}\right)$

${x}^{2} + 6 x + 9 = {\left(x + 3\right)}^{2}$

This will get you

${\left(x + 3\right)}^{2} = 16$

Take the square root from both sides of the equation

$\sqrt{{\left(x + 3\right)}^{2}} = \sqrt{16}$

$x + 3 = \pm 4 \implies {x}_{1 , 2} = - 3 \pm 4 = \left\{\begin{matrix}{x}_{1} = - 3 - 4 = \textcolor{g r e e n}{- 7} \\ {x}_{2} = - 3 + 4 = \textcolor{g r e e n}{1}\end{matrix}\right.$