# How do you solve x^2-12x=-28 by completing the square?

Sep 3, 2016

$x = 8.828 \mathmr{and} x = 3.172$

#### Explanation:

${\left(x \pm p\right)}^{2} = {x}^{2} \pm 2 p x + {p}^{2}$

If you have the correct combination of 3 terms, you can write the expression as ${\left(x \pm p\right)}^{2}$
Add the correct last term each time, to 'complete the square.'

${x}^{2} \textcolor{b l u e}{- 12} x + \square = - 28 \text{ } \square = {\left(\frac{\textcolor{b l u e}{- 12}}{2}\right)}^{2} = \textcolor{red}{36}$
${x}^{2} \textcolor{b l u e}{- 12} x + \textcolor{red}{36} = - 28 + \textcolor{red}{36}$

${\left(x - 6\right)}^{2} = 8 \text{ } - 6 = \frac{\textcolor{b l u e}{- 12}}{2}$

$x - 6 = \pm \sqrt{8}$

$x = \sqrt{8} + 6 = 8.828 \text{ or } x = - \sqrt{8} + 6 = 3.172$