# How do you solve 2x^2-4x+8=0  by completing the square?

Jun 5, 2016

no real solution

#### Explanation:

First, let's divide all terms by 2

$\frac{2}{2} {x}^{2} - \frac{4}{2} x + \frac{8}{2} = 0$
${x}^{2} - 2 x + 4 = 0$

${x}^{2} - 2 x + 1 - 1 + 4 = 0$
so that the initial three terms represent the square of $\left(x - 1\right)$
${\left(x - 1\right)}^{2} + 3 = 0$
${\left(x - 1\right)}^{2} = - 3$