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

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

(^2+3+2)D=4

We move all terms to the left:

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

We multiply parentheses

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

We add all the numbers together, and all the variables

D^2+5D-4=0

a = 1; b = 5; c = -4;
Δ = b2-4ac
Δ = 52-4·1·(-4)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-\sqrt{41}}{2*1}=\frac{-5-\sqrt{41}}{2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{41}}{2*1}=\frac{-5+\sqrt{41}}{2}
$