S=-16t^2+96t+12

Solution for S=-16t^2+96t+12 equation:

=-16S^2+96S+12

We move all terms to the left:

-(-16S^2+96S+12)=0

We get rid of parentheses

16S^2-96S-12=0

a = 16; b = -96; c = -12;
Δ = b2-4ac
Δ = -962-4·16·(-12)
Δ = 9984
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{9984}=\sqrt{256*39}=\sqrt{256}*\sqrt{39}=16\sqrt{39}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-16\sqrt{39}}{2*16}=\frac{96-16\sqrt{39}}{32}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+16\sqrt{39}}{2*16}=\frac{96+16\sqrt{39}}{32}
$