H(t)=-16t^2+92t+24=0

Solution for H(t)=-16t^2+92t+24=0 equation:

(H)=-16H^2+92H+24=0

We move all terms to the left:

(H)-(-16H^2+92H+24)=0

We get rid of parentheses

16H^2-92H+H-24=0

We add all the numbers together, and all the variables

16H^2-91H-24=0

a = 16; b = -91; c = -24;
Δ = b2-4ac
Δ = -912-4·16·(-24)
Δ = 9817
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{-(-91)-\sqrt{9817}}{2*16}=\frac{91-\sqrt{9817}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-91)+\sqrt{9817}}{2*16}=\frac{91+\sqrt{9817}}{32}
$