(D^3+16)y=0

Solution for (D^3+16)y=0 equation:

(^3+16)D=0

We multiply parentheses

D^2+16D=0

a = 1; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·1·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*1}=\frac{-32}{2}
=-16
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*1}=\frac{0}{2}
=0
$