40=5x+4.9x^2

Solution for 40=5x+4.9x^2 equation:

40=5x+4.9x^2

We move all terms to the left:

40-(5x+4.9x^2)=0

We get rid of parentheses

-4.9x^2-5x+40=0

a = -4.9; b = -5; c = +40;
Δ = b2-4ac
Δ = -52-4·(-4.9)·40
Δ = 809
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{-(-5)-\sqrt{809}}{2*-4.9}=\frac{5-\sqrt{809}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{809}}{2*-4.9}=\frac{5+\sqrt{809}}{-9.8}
$