H=-2d^2+12d

Solution for H=-2d^2+12d equation:

=-2H^2+12H

We move all terms to the left:

-(-2H^2+12H)=0

We get rid of parentheses

2H^2-12H=0

a = 2; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·2·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*2}=\frac{0}{4}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*2}=\frac{24}{4}
=6
$