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

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

(^3+3+2)D=0

We multiply parentheses

D^2+3D+2D=0

We add all the numbers together, and all the variables

D^2+5D=0

a = 1; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·1·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*1}=\frac{-10}{2}
=-5
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*1}=\frac{0}{2}
=0
$