S=20t+4.9t^2

Solution for S=20t+4.9t^2 equation:

=20S+4.9S^2

We move all terms to the left:

-(20S+4.9S^2)=0

We get rid of parentheses

-4.9S^2-20S=0

a = -4.9; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·(-4.9)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*-4.9}=\frac{0}{-9.8}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*-4.9}=\frac{40}{-9.8}
=-4+0.8/9.8
$