# How do you solve x^2+30x-7=0 by completing the square?

Aug 19, 2016

$x = 0.2315 \mathmr{and} x = - 30.2315$

#### Explanation:

The method using completing the square is based on:

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

${\left(x - 6\right)}^{2} = {x}^{2} \textcolor{red}{- 12} x + \textcolor{b l u e}{36}$ Note that: ${\left(\frac{\textcolor{red}{- 12}}{2}\right)}^{2} = \textcolor{b l u e}{36}$
This relationship always exists in squaring a binomial.

${x}^{2} + 30 x - 7 = 0 \text{ 7 is not the correct constant}$

Move the 7 to the other side and add in the correct constant on both sides.
${x}^{2} \textcolor{red}{+ 30} x \textcolor{b l u e}{+ 225} = 7 \textcolor{b l u e}{+ 225} \text{ } {\left(\frac{\textcolor{red}{30}}{2}\right)}^{2} = \textcolor{b l u e}{225}$
${\left(x + 15\right)}^{2} = 232 \text{ "x^2 +30x+225" is a square}$

$x + 15 = \pm \sqrt{232} \text{ square root both sides}$

Solve for x twice:

$x = + \sqrt{232} - 15 = 0.2315$

$x = - \sqrt{232} - 15 = - 30.2315$