# How do you solve x^2+5=167?

Feb 5, 2017

$x = \pm 9 \cdot \sqrt{2}$ or $x = \pm 12.728$ to the nearest 3 decimals

#### Explanation:

${x}^{2} + 5 = 167 \text{ }$ there is no $x$ term change to ${x}^{2} = c$

$\therefore {x}^{2} = 162$

$\therefore x = \pm \sqrt{162}$

$\therefore x = \pm \left(\sqrt{3} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{3} \cdot \sqrt{2}\right)$

$\sqrt{3} \cdot \sqrt{3} = 3$

$\therefore x = \pm \left(3 \cdot 3 \cdot \sqrt{2}\right)$

$\therefore x = \pm 9 \sqrt{2}$

Or $x = \pm 9 \cdot 1.414213562$

$\therefore x = \pm 12.728$ to the nearest 3 decimals.