S=-16t^2+52t+3

Solution for S=-16t^2+52t+3 equation:

=-16S^2+52S+3

We move all terms to the left:

-(-16S^2+52S+3)=0

We get rid of parentheses

16S^2-52S-3=0

a = 16; b = -52; c = -3;
Δ = b2-4ac
Δ = -522-4·16·(-3)
Δ = 2896
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:

$\sqrt{\Delta}=\sqrt{2896}=\sqrt{16*181}=\sqrt{16}*\sqrt{181}=4\sqrt{181}$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-4\sqrt{181}}{2*16}=\frac{52-4\sqrt{181}}{32}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+4\sqrt{181}}{2*16}=\frac{52+4\sqrt{181}}{32}
$