S=-16t^2+400t+23000

Solution for S=-16t^2+400t+23000 equation:

=-16S^2+400S+23000

We move all terms to the left:

-(-16S^2+400S+23000)=0

We get rid of parentheses

16S^2-400S-23000=0

a = 16; b = -400; c = -23000;
Δ = b2-4ac
Δ = -4002-4·16·(-23000)
Δ = 1632000
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:

$\sqrt{\Delta}=\sqrt{1632000}=\sqrt{6400*255}=\sqrt{6400}*\sqrt{255}=80\sqrt{255}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-400)-80\sqrt{255}}{2*16}=\frac{400-80\sqrt{255}}{32}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-400)+80\sqrt{255}}{2*16}=\frac{400+80\sqrt{255}}{32}
$