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

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

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

We move all terms to the left:

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

We multiply parentheses

D^2+D^2+4D+4D-3=0

We add all the numbers together, and all the variables

2D^2+8D-3=0

a = 2; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·2·(-3)
Δ = 88
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{22}}{2*2}=\frac{-8-2\sqrt{22}}{4}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{22}}{2*2}=\frac{-8+2\sqrt{22}}{4}
$