V^3+9v^2-36v=0

Solution for V^3+9v^2-36v=0 equation:

^3+9V^2-36V=0

We add all the numbers together, and all the variables

9V^2-36V=0

a = 9; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·9·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*9}=\frac{0}{18}
=0
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*9}=\frac{72}{18}
=4
$