S=-16t^2+108

Solution for S=-16t^2+108 equation:

=-16S^2+108

We move all terms to the left:

-(-16S^2+108)=0

We get rid of parentheses

16S^2-108=0

a = 16; b = 0; c = -108;
Δ = b2-4ac
Δ = 02-4·16·(-108)
Δ = 6912
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:

$\sqrt{\Delta}=\sqrt{6912}=\sqrt{2304*3}=\sqrt{2304}*\sqrt{3}=48\sqrt{3}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{3}}{2*16}=\frac{0-48\sqrt{3}}{32}
=-\frac{48\sqrt{3}}{32}
=-\frac{3\sqrt{3}}{2}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{3}}{2*16}=\frac{0+48\sqrt{3}}{32}
=\frac{48\sqrt{3}}{32}
=\frac{3\sqrt{3}}{2}
$