How do you solve $- 8 - 8 p^{2} = - 31$ $-8-8p^2=-31$ ?

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Answer 1 · 2017-04-24T08:13:11

$p = \pm \sqrt{\frac{23}{8}} ≅ \pm 1.69558$ $p=+-sqrt(23/8) ~= +- 1.69558$

$- 8 - 8 p^{2} = - 31$ $-8-8p^2 = -31$

Multiply through by $- 1$ $-1$

$8 + 8 p^{2} = 31$ $8+8p^2 = 31$

$8 p^{2} = 31 - 8 = 23$ $8p^2 = 31-8 =23$

$p^{2} = \frac{23}{8}$ $p^2 = 23/8$

$p = \pm \sqrt{\frac{23}{8}} ≅ \pm 1.69558$ $p = +- sqrt(23/8) ~=+- 1.69558$

Answer 2 · 2017-04-24T08:15:57

Rearrange your exquation to give $p = \pm \sqrt{\frac{23}{8}}$ $p=+-sqrt(23/8)$

$- 8 - 8 p^{2} = - 31$ $-8-8p^2=-31$
$- 8 p^{2} = - 23$ $-8p^2=-23$
$p^{2} = \frac{23}{8}$ $p^2=23/8$
$p = \pm \sqrt{\frac{23}{8}}$ $p=+-sqrt(23/8)$

How do you solve −8−8p2=−31-8-8p^2=-31?