S=128+112t-16t2

Solution for S=128+112t-16t2 equation:

=128+112S-16S^2

We move all terms to the left:

-(128+112S-16S^2)=0

We get rid of parentheses

16S^2-112S-128=0

a = 16; b = -112; c = -128;
Δ = b2-4ac
Δ = -1122-4·16·(-128)
Δ = 20736
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{20736}=144$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-144}{2*16}=\frac{-32}{32}
=-1
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+144}{2*16}=\frac{256}{32}
=8
$