(D^3+d^2-4d-4)=0

Solution for (D^3+d^2-4d-4)=0 equation:

(^3+D^2-4D-4)=0

We get rid of parentheses

D^2-4D-4+^3=0

We add all the numbers together, and all the variables

D^2-4D=0

a = 1; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·1·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*1}=\frac{0}{2}
=0
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*1}=\frac{8}{2}
=4
$