6d^2+2d=20

Solution for 6d^2+2d=20 equation:

6d^2+2d=20

We move all terms to the left:

6d^2+2d-(20)=0

a = 6; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·6·(-20)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*6}=\frac{-24}{12}
=-2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*6}=\frac{20}{12}
=1+2/3
$