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

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

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

We multiply parentheses

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

We add all the numbers together, and all the variables

2D^2+4D=0

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

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