(D^3-D^2-8D-12)y=0

Solution for (D^3-D^2-8D-12)y=0 equation:

(^3-^2-8-12)D=0

We multiply parentheses

D^2+D^2-8D-12D=0

We add all the numbers together, and all the variables

2D^2-20D=0

a = 2; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·2·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*2}=\frac{0}{4}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*2}=\frac{40}{4}
=10
$