(D^3+5D)y=15

Solution for (D^3+5D)y=15 equation:

(^3+5)D=15

We move all terms to the left:

(^3+5)D-(15)=0

We multiply parentheses

D^2+5D-15=0

a = 1; b = 5; c = -15;
Δ = b2-4ac
Δ = 52-4·1·(-15)
Δ = 85
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{-(5)-\sqrt{85}}{2*1}=\frac{-5-\sqrt{85}}{2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{85}}{2*1}=\frac{-5+\sqrt{85}}{2}
$