# How do you solve x^2 - 8x - 6=0 using completing the square?

Jul 23, 2015

First you change the form into ${x}^{2} - 8 x = 6$
(add $6$ to both sides)

#### Explanation:

Method: you halve the $x$-coefficient ($- 8$) to get $- 4$. The square of this is to be added to both sides.

$\to {x}^{2} - 8 x + 16 = 6 + 16$

${\left(x - 4\right)}^{2} = 22 \to x - 4 = \pm \sqrt{22}$

${x}_{1 , 2} = 4 \pm \sqrt{22}$ (two solutions)
graph{x^2-8x-6 [-52.02, 52.07, -26, 26]}