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

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

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

We multiply parentheses

D^2+4D+4D=0

We add all the numbers together, and all the variables

D^2+8D=0

a = 1; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·1·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*1}=\frac{-16}{2}
=-8
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*1}=\frac{0}{2}
=0
$