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

Solution for H=-16t^2+24t+40=0 equation:

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

We move all terms to the left:

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

We get rid of parentheses

16H^2-24H-40=0

a = 16; b = -24; c = -40;
Δ = b2-4ac
Δ = -242-4·16·(-40)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-56}{2*16}=\frac{-32}{32}
=-1
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+56}{2*16}=\frac{80}{32}
=2+1/2
$