S(t)=-16t^2+200t+4

Solution for S(t)=-16t^2+200t+4 equation:

(S)=-16S^2+200S+4

We move all terms to the left:

(S)-(-16S^2+200S+4)=0

We get rid of parentheses

16S^2-200S+S-4=0

We add all the numbers together, and all the variables

16S^2-199S-4=0

a = 16; b = -199; c = -4;
Δ = b2-4ac
Δ = -1992-4·16·(-4)
Δ = 39857
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-199)-\sqrt{39857}}{2*16}=\frac{199-\sqrt{39857}}{32}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-199)+\sqrt{39857}}{2*16}=\frac{199+\sqrt{39857}}{32}
$