H=-0.5d^2+2d+6

Solution for H=-0.5d^2+2d+6 equation:

=-0.5H^2+2H+6

We move all terms to the left:

-(-0.5H^2+2H+6)=0

We get rid of parentheses

0.5H^2-2H-6=0

a = 0.5; b = -2; c = -6;
Δ = b2-4ac
Δ = -22-4·0.5·(-6)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*0.5}=\frac{-2}{1}
=-2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*0.5}=\frac{6}{1}
=6
$