# How do you solve 2x^2 -12x + 11=0 by completing the square?

##### 1 Answer
May 13, 2015

First divide through by 2 to get:

${x}^{2} - 6 x + \frac{11}{2} = 0$

Now ${\left(x - 3\right)}^{2}$ = ${x}^{2} - 6 x + 9$

So we can write

$0 = {x}^{2} - 6 x + \frac{11}{2}$

$= {x}^{2} - 6 x + 9 - 9 + \frac{11}{2}$

$= {\left(x - 3\right)}^{2} - 9 + \frac{11}{2}$

$= {\left(x - 3\right)}^{2} - \frac{18}{2} + \frac{11}{2}$

$= {\left(x - 3\right)}^{2} - \frac{7}{2}$

Adding $\frac{7}{2}$ to both sides we get

${\left(x - 3\right)}^{2} = \frac{7}{2}$

So $\left(x - 3\right) = \pm \sqrt{\frac{7}{2}}$

Add 3 to both sides to get

$x = 3 \pm \sqrt{\frac{7}{2}} = 3 \pm \frac{\sqrt{7}}{\sqrt{2}}$

If you prefer, $\sqrt{\frac{7}{2}} = \sqrt{\frac{14}{4}} = \frac{\sqrt{14}}{2}$, so

$x = 3 \pm \frac{\sqrt{14}}{2}$

In general,

$a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

Hence $a {x}^{2} + b x + c = 0$ has solutions

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$