(D^3+9D)y=27

Solution for (D^3+9D)y=27 equation:

(^3+9)D=27

We move all terms to the left:

(^3+9)D-(27)=0

We multiply parentheses

D^2+9D-27=0

a = 1; b = 9; c = -27;
Δ = b2-4ac
Δ = 92-4·1·(-27)
Δ = 189
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{21}}{2*1}=\frac{-9-3\sqrt{21}}{2}
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{21}}{2*1}=\frac{-9+3\sqrt{21}}{2}
$