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