H(t)=-16t2+96t+6

Solution for H(t)=-16t2+96t+6 equation:

(H)=-16H^2+96H+6

We move all terms to the left:

(H)-(-16H^2+96H+6)=0

We get rid of parentheses

16H^2-96H+H-6=0

We add all the numbers together, and all the variables

16H^2-95H-6=0

a = 16; b = -95; c = -6;
Δ = b2-4ac
Δ = -952-4·16·(-6)
Δ = 9409
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9409}=97$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-97}{2*16}=\frac{-2}{32}
=-1/16
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+97}{2*16}=\frac{192}{32}
=6
$