# How do you solve x^2-12x+20=0 by completing the square?

Aug 12, 2015

${x}_{1 , 2} = 6 \pm 4$

#### Explanation:

$\textcolor{b l u e}{{x}^{2} + \frac{b}{a} x = - \frac{c}{a}}$

This can be done by adding $- 20$ to both sides of the equation

${x}^{2} - 12 x + \textcolor{red}{\cancel{\textcolor{b l a c k}{20}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{20}}} = 0 - 20$

${x}^{2} - 12 x = - 20$

Now use the coefficient of the $x$-term to find a term that, when added to both sides of the equation, will allow you to write the left side as the square of a binomial.

More specifically, you need to divide this coefficient by $2$ and square the result.

${\left(\frac{- 12}{2}\right)}^{2} = {\left(- 6\right)}^{2} = 36$

So, add $36$ to both sides of the equation to get

${x}^{2} - 12 x + 36 = - 20 + 36$

The left side of the equation can now be written as

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

This means that you have

${\left(x - 6\right)}^{2} = 16$

Take the square root of both sides of the equation to get

$\sqrt{{\left(x - 6\right)}^{2}} = \sqrt{16}$

$x - 6 = \pm 4$

$x = 6 \pm 4 = \left\{\begin{matrix}{x}_{1} = 6 + 4 = \textcolor{g r e e n}{10} \\ {x}_{2} = 6 - 4 = \textcolor{g r e e n}{2}\end{matrix}\right.$