H(t)=-16t^2+40t+72

Solution for H(t)=-16t^2+40t+72 equation:

(H)=-16H^2+40H+72

We move all terms to the left:

(H)-(-16H^2+40H+72)=0

We get rid of parentheses

16H^2-40H+H-72=0

We add all the numbers together, and all the variables

16H^2-39H-72=0

a = 16; b = -39; c = -72;
Δ = b2-4ac
Δ = -392-4·16·(-72)
Δ = 6129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{6129}=\sqrt{9*681}=\sqrt{9}*\sqrt{681}=3\sqrt{681}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-3\sqrt{681}}{2*16}=\frac{39-3\sqrt{681}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+3\sqrt{681}}{2*16}=\frac{39+3\sqrt{681}}{32}
$