S(t)=-16t^2+10t+4

Solution for S(t)=-16t^2+10t+4 equation:

(S)=-16S^2+10S+4

We move all terms to the left:

(S)-(-16S^2+10S+4)=0

We get rid of parentheses

16S^2-10S+S-4=0

We add all the numbers together, and all the variables

16S^2-9S-4=0

a = 16; b = -9; c = -4;
Δ = b2-4ac
Δ = -92-4·16·(-4)
Δ = 337
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{337}}{2*16}=\frac{9-\sqrt{337}}{32}
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{337}}{2*16}=\frac{9+\sqrt{337}}{32}
$