H=H+15t-5t^2

Solution for H=H+15t-5t^2 equation:

=+15H-5H^2

We move all terms to the left:

-(+15H-5H^2)=0

We get rid of parentheses

5H^2-15H=0

a = 5; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·5·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*5}=\frac{0}{10}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*5}=\frac{30}{10}
=3
$