(D^2+3D)y=5

Solution for (D^2+3D)y=5 equation:

(^2+3)D=5

We move all terms to the left:

(^2+3)D-(5)=0

We multiply parentheses

D^2+3D-5=0

a = 1; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·1·(-5)
Δ = 29
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{-(3)-\sqrt{29}}{2*1}=\frac{-3-\sqrt{29}}{2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{29}}{2*1}=\frac{-3+\sqrt{29}}{2}
$