300=19x+4.9x^2

Solution for 300=19x+4.9x^2 equation:

300=19x+4.9x^2

We move all terms to the left:

300-(19x+4.9x^2)=0

We get rid of parentheses

-4.9x^2-19x+300=0

a = -4.9; b = -19; c = +300;
Δ = b2-4ac
Δ = -192-4·(-4.9)·300
Δ = 6241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6241}=79$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-79}{2*-4.9}=\frac{-60}{-9.8}
=6+1/8.16666666667
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+79}{2*-4.9}=\frac{98}{-9.8}
=-10
$