# How do you solve the equation 6x^2-5x-13=x^2-11 by completing the square?

Aug 20, 2017

$x = \sqrt{\frac{13}{20}} + \frac{1}{2}$
$x = - \sqrt{\frac{13}{20}} + \frac{1}{2}$

#### Explanation:

Given -

$6 {x}^{2} - 5 x - 13 = {x}^{2} - 11$

Take all the terms to the left-hand side

$6 {x}^{2} - 5 x - 13 - {x}^{2} + 11 = 0$

Simplify it.

$5 {x}^{2} - 5 x - 2 = 0$

Take the constant term to the right-hand side

$5 {x}^{2} - 5 x = 2$

Divide all the terms by the coefficient of ${x}^{2}$

$\frac{5 {x}^{2}}{5} - \frac{5 x}{5} = \frac{2}{5}$

${x}^{2} - x = \frac{2}{5}$

Take half the coefficient of $x$, square it and add it to both sides

${x}^{2} - x + \frac{1}{4} = \frac{2}{5} + \frac{1}{4} = \frac{8 + 5}{20} = \frac{13}{20}$

${\left(x - \frac{1}{2}\right)}^{2} = \frac{13}{20}$

$\left(x - \frac{1}{2}\right) = \pm \sqrt{\frac{13}{20}}$

$x = \sqrt{\frac{13}{20}} + \frac{1}{2}$
$x = - \sqrt{\frac{13}{20}} + \frac{1}{2}$