H(t)=-16t^2+20t+2.5

Solution for H(t)=-16t^2+20t+2.5 equation:

(H)=-16H^2+20H+2.5

We move all terms to the left:

(H)-(-16H^2+20H+2.5)=0

We get rid of parentheses

16H^2-20H+H-2.5=0

We add all the numbers together, and all the variables

16H^2-19H-2.5=0

a = 16; b = -19; c = -2.5;
Δ = b2-4ac
Δ = -192-4·16·(-2.5)
Δ = 521
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{-(-19)-\sqrt{521}}{2*16}=\frac{19-\sqrt{521}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{521}}{2*16}=\frac{19+\sqrt{521}}{32}
$