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

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

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

We move all terms to the left:

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

We get rid of parentheses

16S^2-96S+S=0

We add all the numbers together, and all the variables

16S^2-95S=0

a = 16; b = -95; c = 0;
Δ = b2-4ac
Δ = -952-4·16·0
Δ = 9025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9025}=95$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-95)-95}{2*16}=\frac{0}{32}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-95)+95}{2*16}=\frac{190}{32}
=5+15/16
$