How do you solve the quadratic using the quadratic formula given $v^{2} + 2 v - 8 = 0$ $v^2+2v-8=0$ ?

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Answer 1 · 2017-07-09T04:35:01

v = -4 or 2

Starting with the quadratic formula:

$v = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $v = (-b\pm sqrt(b^2-4ac))/(2a)$

where

$a = 1, b = 2, and c = - 8$ $a=1, b=2, and c=-8$

Gives:

$v = \frac{- 2 \pm \sqrt{2^{2} - 4 \cdot 1 \cdot (- 8)}}{2 \cdot 1}$ $v= (-2\pm sqrt(2^2-4*1*(-8)))/(2*1)$

$v = \frac{- 2 \pm \sqrt{4 + 32}}{2}$ $v= (-2\pm sqrt(4+32))/2$

$v = \frac{- 2 \pm \sqrt{36}}{2}$ $v= (-2\pm sqrt(36))/2$

$v = \frac{- 2 \pm 6}{2}$ $v= (-2\pm 6)/2$

$v = - \frac{8}{2} or \frac{4}{2}$ $v= -8/2 or 4/2$

$v = - 4 or 2$ $v = -4 or 2$

How do you solve the quadratic using the quadratic formula given v2+2v−8=0v^2+2v-8=0?