H(t)=-16t^2+96t+48

Solution for H(t)=-16t^2+96t+48 equation:

(H)=-16H^2+96H+48

We move all terms to the left:

(H)-(-16H^2+96H+48)=0

We get rid of parentheses

16H^2-96H+H-48=0

We add all the numbers together, and all the variables

16H^2-95H-48=0

a = 16; b = -95; c = -48;
Δ = b2-4ac
Δ = -952-4·16·(-48)
Δ = 12097
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{-(-95)-\sqrt{12097}}{2*16}=\frac{95-\sqrt{12097}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+\sqrt{12097}}{2*16}=\frac{95+\sqrt{12097}}{32}
$