S(3200)=16t^2

Solution for S(3200)=16t^2 equation:

(3200)=16S^2

We move all terms to the left:

(3200)-(16S^2)=0

a = -16; b = 0; c = +3200;
Δ = b2-4ac
Δ = 02-4·(-16)·3200
Δ = 204800
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:

$\sqrt{\Delta}=\sqrt{204800}=\sqrt{102400*2}=\sqrt{102400}*\sqrt{2}=320\sqrt{2}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-320\sqrt{2}}{2*-16}=\frac{0-320\sqrt{2}}{-32}
=-\frac{320\sqrt{2}}{-32}
=-\frac{10\sqrt{2}}{-1}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+320\sqrt{2}}{2*-16}=\frac{0+320\sqrt{2}}{-32}
=\frac{320\sqrt{2}}{-32}
=\frac{10\sqrt{2}}{-1}
$