H(t)=-16t^2+32t+108

Solution for H(t)=-16t^2+32t+108 equation:

(H)=-16H^2+32H+108

We move all terms to the left:

(H)-(-16H^2+32H+108)=0

We get rid of parentheses

16H^2-32H+H-108=0

We add all the numbers together, and all the variables

16H^2-31H-108=0

a = 16; b = -31; c = -108;
Δ = b2-4ac
Δ = -312-4·16·(-108)
Δ = 7873
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{-(-31)-\sqrt{7873}}{2*16}=\frac{31-\sqrt{7873}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{7873}}{2*16}=\frac{31+\sqrt{7873}}{32}
$