20v^2-16v-16=0

Solution for 20v^2-16v-16=0 equation:

20v^2-16v-16=0

a = 20; b = -16; c = -16;
Δ = b2-4ac
Δ = -162-4·20·(-16)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{6}}{2*20}=\frac{16-16\sqrt{6}}{40}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{6}}{2*20}=\frac{16+16\sqrt{6}}{40}
$