# How do you solve x^2 - 8x + 3 = 0?

Apr 30, 2016

$x = 4 \pm \sqrt{13}$

#### Explanation:

Complete the square and use the difference of squares identity:

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

with $a = \left(x - 4\right)$ and $b = \sqrt{13}$ as follows:

$0 = {x}^{2} - 8 x + 3$

$= {\left(x - 4\right)}^{2} - 16 + 3$

$= {\left(x - 4\right)}^{2} - 13$

$= {\left(x - 4\right)}^{2} - {\left(\sqrt{13}\right)}^{2}$

$= \left(\left(x - 4\right) - \sqrt{13}\right) \left(\left(x - 4\right) + \sqrt{13}\right)$

$= \left(x - 4 - \sqrt{13}\right) \left(x - 4 + \sqrt{13}\right)$

So:

$x = 4 \pm \sqrt{13}$