(D^2+4D+5)y=5

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

(^2+4+5)D=5

We move all terms to the left:

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

We multiply parentheses

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

We add all the numbers together, and all the variables

D^2+9D-5=0

a = 1; b = 9; c = -5;
Δ = b2-4ac
Δ = 92-4·1·(-5)
Δ = 101
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{-(9)-\sqrt{101}}{2*1}=\frac{-9-\sqrt{101}}{2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{101}}{2*1}=\frac{-9+\sqrt{101}}{2}
$