0=-4.9x^2+98x

Solution for 0=-4.9x^2+98x equation:

0=-4.9x^2+98x

We move all terms to the left:

0-(-4.9x^2+98x)=0

We add all the numbers together, and all the variables

-(-4.9x^2+98x)=0

We get rid of parentheses

4.9x^2-98x=0

a = 4.9; b = -98; c = 0;
Δ = b2-4ac
Δ = -982-4·4.9·0
Δ = 9604
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9604}=98$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-98)-98}{2*4.9}=\frac{0}{9.8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-98)+98}{2*4.9}=\frac{196}{9.8}
=20
$