300=10x+4.905x^2

Solution for 300=10x+4.905x^2 equation:

300=10x+4.905x^2

We move all terms to the left:

300-(10x+4.905x^2)=0

We get rid of parentheses

-4.905x^2-10x+300=0

a = -4.905; b = -10; c = +300;
Δ = b2-4ac
Δ = -102-4·(-4.905)·300
Δ = 5986
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{-(-10)-\sqrt{5986}}{2*-4.905}=\frac{10-\sqrt{5986}}{-9.81}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{5986}}{2*-4.905}=\frac{10+\sqrt{5986}}{-9.81}
$