S^3+4s^2+4s+4=0

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

^3+4S^2+4S+4=0

We add all the numbers together, and all the variables

4S^2+4S=0

a = 4; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·4·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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