H=-16t^2+45t+20

Solution for H=-16t^2+45t+20 equation:

=-16H^2+45H+20

We move all terms to the left:

-(-16H^2+45H+20)=0

We get rid of parentheses

16H^2-45H-20=0

a = 16; b = -45; c = -20;
Δ = b2-4ac
Δ = -452-4·16·(-20)
Δ = 3305
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{-(-45)-\sqrt{3305}}{2*16}=\frac{45-\sqrt{3305}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+\sqrt{3305}}{2*16}=\frac{45+\sqrt{3305}}{32}
$