S=16t^2+112t

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

=16S^2+112S

We move all terms to the left:

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

We get rid of parentheses

-16S^2-112S=0

a = -16; b = -112; c = 0;
Δ = b2-4ac
Δ = -1122-4·(-16)·0
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-112}{2*-16}=\frac{0}{-32}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+112}{2*-16}=\frac{224}{-32}
=-7
$