H=-16t^2+15t+20

Solution for H=-16t^2+15t+20 equation:

=-16H^2+15H+20

We move all terms to the left:

-(-16H^2+15H+20)=0

We get rid of parentheses

16H^2-15H-20=0

a = 16; b = -15; c = -20;
Δ = b2-4ac
Δ = -152-4·16·(-20)
Δ = 1505
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{-(-15)-\sqrt{1505}}{2*16}=\frac{15-\sqrt{1505}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{1505}}{2*16}=\frac{15+\sqrt{1505}}{32}
$