(D^2+4D)y=5

Solution for (D^2+4D)y=5 equation:

(^2+4)D=5

We move all terms to the left:

(^2+4)D-(5)=0

We multiply parentheses

D^2+4D-5=0

a = 1; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·1·(-5)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-6}{2*1}=\frac{-10}{2}
=-5
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+6}{2*1}=\frac{2}{2}
=1
$