H(t)=-16t^2-30t+100

Solution for H(t)=-16t^2-30t+100 equation:

(H)=-16H^2-30H+100

We move all terms to the left:

(H)-(-16H^2-30H+100)=0

We get rid of parentheses

16H^2+30H+H-100=0

We add all the numbers together, and all the variables

16H^2+31H-100=0

a = 16; b = 31; c = -100;
Δ = b2-4ac
Δ = 312-4·16·(-100)
Δ = 7361
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{7361}}{2*16}=\frac{-31-\sqrt{7361}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{7361}}{2*16}=\frac{-31+\sqrt{7361}}{32}
$