S=-16t^2+112t+4

Solution for S=-16t^2+112t+4 equation:

=-16S^2+112S+4

We move all terms to the left:

-(-16S^2+112S+4)=0

We get rid of parentheses

16S^2-112S-4=0

a = 16; b = -112; c = -4;
Δ = b2-4ac
Δ = -1122-4·16·(-4)
Δ = 12800
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:

$\sqrt{\Delta}=\sqrt{12800}=\sqrt{6400*2}=\sqrt{6400}*\sqrt{2}=80\sqrt{2}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-80\sqrt{2}}{2*16}=\frac{112-80\sqrt{2}}{32}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+80\sqrt{2}}{2*16}=\frac{112+80\sqrt{2}}{32}
$