# How do you solve 3x^2-4x-2=0 by completing the square?

Oct 24, 2016

$x = \frac{2}{3} + \frac{\sqrt{10}}{3}$ , $x = \frac{2}{3} - \frac{\sqrt{10}}{3}$

#### Explanation:

Firstly, we need to deal with the coefficient of ${x}^{2}$ (the 3).

In order to remove the 3 we divide the whole expression by 3.

$\frac{3 {x}^{2} - 4 x - 2}{3} = \frac{0}{3}$

${x}^{2} - \frac{4}{3} x - \frac{2}{3} = 0$

next, we complete the square, here is the general formula

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

so using this formula,

${x}^{2} - \frac{4}{3} x - \frac{2}{3} = 0$

goes to

${\left(x - \frac{4}{6}\right)}^{2} - \frac{16}{36} - \frac{2}{3} = 0$

simplifying down,

${\left(x - \frac{2}{3}\right)}^{2} - \frac{10}{9} = 0$

rearrange for x,

${\left(x - \frac{2}{3}\right)}^{2} = \frac{10}{9}$

$x - \frac{2}{3} = \pm \sqrt{\frac{10}{9}}$

$x = \frac{2}{3} + \frac{\sqrt{10}}{3}$ , $x = \frac{2}{3} - \frac{\sqrt{10}}{3}$

these are the two solutions.