# How do you solve  x^2 - 6x - 3 = 0 ?

Aug 31, 2016

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

#### Explanation:

Complete the square then use the difference of squares identity:

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

with $a = \left(x - 3\right)$ and $b = 2 \sqrt{3}$ as follows:

$0 = {x}^{2} - 6 x - 3$

$\textcolor{w h i t e}{0} = {x}^{2} - 6 x + 9 - 12$

$\textcolor{w h i t e}{0} = {\left(x - 3\right)}^{2} - {\left(2 \sqrt{3}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(x - 3\right) - 2 \sqrt{3}\right) \left(\left(x - 3\right) + 2 \sqrt{3}\right)$

$\textcolor{w h i t e}{0} = \left(x - 3 - 2 \sqrt{3}\right) \left(x - 3 + 2 \sqrt{3}\right)$

Hence:

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