(D^3+D^2+4D+4)*y=0

Solution for (D^3+D^2+4D+4)*y=0 equation:

(^3+^2+4+4)*D=0

We multiply parentheses

D^2+D^2+4D+4D=0

We add all the numbers together, and all the variables

2D^2+8D=0

a = 2; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*2}=\frac{-16}{4}
=-4
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*2}=\frac{0}{4}
=0
$