v^2+20v-40=-8

Solution for v^2+20v-40=-8 equation:

v^2+20v-40=-8

We move all terms to the left:

v^2+20v-40-(-8)=0

We add all the numbers together, and all the variables

v^2+20v-32=0

a = 1; b = 20; c = -32;
Δ = b2-4ac
Δ = 202-4·1·(-32)
Δ = 528
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{33}}{2*1}=\frac{-20-4\sqrt{33}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{33}}{2*1}=\frac{-20+4\sqrt{33}}{2}
$