# How do you solve 2x^2 - 6x +1 = 0 by completing the square?

Jan 21, 2017

$x = \sqrt{\frac{7}{4}} + \frac{3}{2}$

$x = - \sqrt{\frac{7}{4}} + \frac{3}{2}$

#### Explanation:

Given -

$2 {x}^{2} - 6 x + 1 = 0$

$2 {x}^{2} - 6 x = - 1$

$\frac{2 {x}^{2}}{2} - \frac{6 x}{2} = \frac{- 1}{2}$

${x}^{2} - 3 x = \frac{- 1}{2}$

${x}^{2} - 3 x + \frac{9}{4} = \frac{- 1}{2} + \frac{9}{4}$

${\left(x - \frac{3}{2}\right)}^{2} = \frac{- 2 + 9}{4} = \frac{7}{4}$

$\left(x - \frac{3}{2}\right) = \pm \sqrt{\frac{7}{4}}$

$x = \sqrt{\frac{7}{4}} + \frac{3}{2}$

$x = - \sqrt{\frac{7}{4}} + \frac{3}{2}$