S^3-4s^2-12s=0

Solution for S^3-4s^2-12s=0 equation:

^3-4S^2-12S=0

We add all the numbers together, and all the variables

-4S^2-12S=0

a = -4; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·(-4)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*-4}=\frac{0}{-8}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*-4}=\frac{24}{-8}
=-3
$