V^3-2v^2-16v-32=0

Solution for V^3-2v^2-16v-32=0 equation:

^3-2V^2-16V-32=0

We add all the numbers together, and all the variables

-2V^2-16V=0

a = -2; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·(-2)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*-2}=\frac{0}{-4}
=0
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-2}=\frac{32}{-4}
=-8
$