S=120+96t-16t^2

Solution for S=120+96t-16t^2 equation:

=120+96S-16S^2

We move all terms to the left:

-(120+96S-16S^2)=0

We get rid of parentheses

16S^2-96S-120=0

a = 16; b = -96; c = -120;
Δ = b2-4ac
Δ = -962-4·16·(-120)
Δ = 16896
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:

$\sqrt{\Delta}=\sqrt{16896}=\sqrt{256*66}=\sqrt{256}*\sqrt{66}=16\sqrt{66}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-16\sqrt{66}}{2*16}=\frac{96-16\sqrt{66}}{32}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+16\sqrt{66}}{2*16}=\frac{96+16\sqrt{66}}{32}
$