(D^3+D^2+D+1)y=0

Solution for (D^3+D^2+D+1)y=0 equation:

(^3+^2++1)D=0

We use the square of the difference formula

(^3+^2+1)D=0

We multiply parentheses

D^2+D^2+D=0

We add all the numbers together, and all the variables

2D^2+D=0

a = 2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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