S(t)=-20t^2+200t

Solution for S(t)=-20t^2+200t equation:

(S)=-20S^2+200S

We move all terms to the left:

(S)-(-20S^2+200S)=0

We get rid of parentheses

20S^2-200S+S=0

We add all the numbers together, and all the variables

20S^2-199S=0

a = 20; b = -199; c = 0;
Δ = b2-4ac
Δ = -1992-4·20·0
Δ = 39601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{39601}=199$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-199)-199}{2*20}=\frac{0}{40}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-199)+199}{2*20}=\frac{398}{40}
=9+19/20
$