S=-16t^2+300

Solution for S=-16t^2+300 equation:

=-16S^2+300

We move all terms to the left:

-(-16S^2+300)=0

We get rid of parentheses

16S^2-300=0

a = 16; b = 0; c = -300;
Δ = b2-4ac
Δ = 02-4·16·(-300)
Δ = 19200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{19200}=\sqrt{6400*3}=\sqrt{6400}*\sqrt{3}=80\sqrt{3}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-80\sqrt{3}}{2*16}=\frac{0-80\sqrt{3}}{32}
=-\frac{80\sqrt{3}}{32}
=-\frac{5\sqrt{3}}{2}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+80\sqrt{3}}{2*16}=\frac{0+80\sqrt{3}}{32}
=\frac{80\sqrt{3}}{32}
=\frac{5\sqrt{3}}{2}
$