H(t)=-16t^2+40t+15

Solution for H(t)=-16t^2+40t+15 equation:

(H)=-16H^2+40H+15

We move all terms to the left:

(H)-(-16H^2+40H+15)=0

We get rid of parentheses

16H^2-40H+H-15=0

We add all the numbers together, and all the variables

16H^2-39H-15=0

a = 16; b = -39; c = -15;
Δ = b2-4ac
Δ = -392-4·16·(-15)
Δ = 2481
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{-(-39)-\sqrt{2481}}{2*16}=\frac{39-\sqrt{2481}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{2481}}{2*16}=\frac{39+\sqrt{2481}}{32}
$