0=50+20x-4.905x^2

Solution for 0=50+20x-4.905x^2 equation:

0=50+20x-4.905x^2

We move all terms to the left:

0-(50+20x-4.905x^2)=0

We add all the numbers together, and all the variables

-(50+20x-4.905x^2)=0

We get rid of parentheses

4.905x^2-20x-50=0

a = 4.905; b = -20; c = -50;
Δ = b2-4ac
Δ = -202-4·4.905·(-50)
Δ = 1381
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-\sqrt{1381}}{2*4.905}=\frac{20-\sqrt{1381}}{9.81}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{1381}}{2*4.905}=\frac{20+\sqrt{1381}}{9.81}
$