S=-4.9t^2+49t

Solution for S=-4.9t^2+49t equation:

=-4.9S^2+49S

We move all terms to the left:

-(-4.9S^2+49S)=0

We get rid of parentheses

4.9S^2-49S=0

a = 4.9; b = -49; c = 0;
Δ = b2-4ac
Δ = -492-4·4.9·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$S_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-49)-49}{2*4.9}=\frac{0}{9.8}
=0
$
$S_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+49}{2*4.9}=\frac{98}{9.8}
=10
$