How do you solve by completing the square: x^2 + 6x - 5 = 0?

Mar 4, 2018

$x = - 3 + \sqrt{14} \mathmr{and} x = - 3 - \sqrt{14}$

Explanation:

${x}^{2} + 6 x - 5 = 0$
$\Rightarrow {x}^{2} + 6 x + \textcolor{red}{9 - 14} = 0$
$\Rightarrow \left({x}^{2} + 6 x + 9\right) = 14$
$\Rightarrow \textcolor{b l u e}{{\left(x + 3\right)}^{2} = {\left(\sqrt{14}\right)}^{2}}$
$\Rightarrow x + 3 = \pm \sqrt{14}$
$\Rightarrow x + 3 = \sqrt{14} \mathmr{and} x + 3 = - \sqrt{14}$
$\Rightarrow x = - 3 + \sqrt{14} \mathmr{and} x = - 3 - \sqrt{14}$