50=2x+4.905x^2

Solution for 50=2x+4.905x^2 equation:

50=2x+4.905x^2

We move all terms to the left:

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

We get rid of parentheses

-4.905x^2-2x+50=0

a = -4.905; b = -2; c = +50;
Δ = b2-4ac
Δ = -22-4·(-4.905)·50
Δ = 985
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{-(-2)-\sqrt{985}}{2*-4.905}=\frac{2-\sqrt{985}}{-9.81}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+\sqrt{985}}{2*-4.905}=\frac{2+\sqrt{985}}{-9.81}
$