H(t)=-16t2+24t+40

Solution for H(t)=-16t2+24t+40 equation:

(H)=-16H^2+24H+40

We move all terms to the left:

(H)-(-16H^2+24H+40)=0

We get rid of parentheses

16H^2-24H+H-40=0

We add all the numbers together, and all the variables

16H^2-23H-40=0

a = 16; b = -23; c = -40;
Δ = b2-4ac
Δ = -232-4·16·(-40)
Δ = 3089
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{-(-23)-\sqrt{3089}}{2*16}=\frac{23-\sqrt{3089}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{3089}}{2*16}=\frac{23+\sqrt{3089}}{32}
$