H(t)=-16t^2+96t+2

Solution for H(t)=-16t^2+96t+2 equation:

(H)=-16H^2+96H+2

We move all terms to the left:

(H)-(-16H^2+96H+2)=0

We get rid of parentheses

16H^2-96H+H-2=0

We add all the numbers together, and all the variables

16H^2-95H-2=0

a = 16; b = -95; c = -2;
Δ = b2-4ac
Δ = -952-4·16·(-2)
Δ = 9153
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9153}=\sqrt{81*113}=\sqrt{81}*\sqrt{113}=9\sqrt{113}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-9\sqrt{113}}{2*16}=\frac{95-9\sqrt{113}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+9\sqrt{113}}{2*16}=\frac{95+9\sqrt{113}}{32}
$