H=0.01x^2+0.9x

Solution for H=0.01x^2+0.9x equation:

=0.01H^2+0.9H

We move all terms to the left:

-(0.01H^2+0.9H)=0

We get rid of parentheses

-0.01H^2-0.9H=0

a = -0.01; b = -0.9; c = 0;
Δ = b2-4ac
Δ = -0.92-4·(-0.01)·0
Δ = 0.81
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{-(-0.9)-\sqrt{0.81}}{2*-0.01}=\frac{0.9-\sqrt{0.81}}{-0.02}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.9)+\sqrt{0.81}}{2*-0.01}=\frac{0.9+\sqrt{0.81}}{-0.02}
$