H=-0.02d^2+1.2d+4

Solution for H=-0.02d^2+1.2d+4 equation:

=-0.02H^2+1.2H+4

We move all terms to the left:

-(-0.02H^2+1.2H+4)=0

We get rid of parentheses

0.02H^2-1.2H-4=0

a = 0.02; b = -1.2; c = -4;
Δ = b2-4ac
Δ = -1.22-4·0.02·(-4)
Δ = 1.76
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1.2)-\sqrt{1.76}}{2*0.02}=\frac{1.2-\sqrt{1.76}}{0.04}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1.2)+\sqrt{1.76}}{2*0.02}=\frac{1.2+\sqrt{1.76}}{0.04}
$