# How do you solve 6-p^2/8=-4?

Oct 30, 2016

$p = \pm 4 \sqrt{5}$

#### Explanation:

$6 - {p}^{2} / 8 = - 4$

$\Leftrightarrow 6 + 4 = {p}^{2} / 8$

or ${p}^{2} / 8 = 10$

or ${p}^{2} = 80$

or ${p}^{2} - 80 = 0$

or ${p}^{2} - {\left(\sqrt{80}\right)}^{2} = 0$

or $\left(p + \sqrt{80}\right) \left(p - \sqrt{80}\right) = 0$

i.e. either $p + \sqrt{80} = 0$ or $p - \sqrt{80} = 0$

i.e. either $p = - \sqrt{80}$ or $p = \sqrt{80}$

i.e. p=+-sqrt(4×4×5)=+-4sqrt5