S(t)=-16(t)^2+S0

Solution for S(t)=-16(t)^2+S0 equation:

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

We move all terms to the left:

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

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

We get rid of parentheses

16S^2+S-0=0

We add all the numbers together, and all the variables

16S^2+S=0

a = 16; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·16·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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