S=-16t^2+98t

Solution for S=-16t^2+98t equation:

=-16S^2+98S

We move all terms to the left:

-(-16S^2+98S)=0

We get rid of parentheses

16S^2-98S=0

a = 16; b = -98; c = 0;
Δ = b2-4ac
Δ = -982-4·16·0
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-98}{2*16}=\frac{0}{32}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+98}{2*16}=\frac{196}{32}
=6+1/8
$