(D^2+4D+20)y=0

Solution for (D^2+4D+20)y=0 equation:

(^2+4+20)D=0

We multiply parentheses

D^2+4D+20D=0

We add all the numbers together, and all the variables

D^2+24D=0

a = 1; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·1·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*1}=\frac{-48}{2}
=-24
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*1}=\frac{0}{2}
=0
$