D^3+6d^2+11d+6=0

Solution for D^3+6d^2+11d+6=0 equation:

^3+6D^2+11D+6=0

We add all the numbers together, and all the variables

6D^2+11D=0

a = 6; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·6·0
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*6}=\frac{-22}{12}
=-1+5/6
$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*6}=\frac{0}{12}
=0
$