D^3-4d^2+d+6=0

Solution for D^3-4d^2+d+6=0 equation:

^3-4D^2+D+6=0

We add all the numbers together, and all the variables

-4D^2+D=0

a = -4; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-4)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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