45=21x+4.9x^2

Solution for 45=21x+4.9x^2 equation:

45=21x+4.9x^2

We move all terms to the left:

45-(21x+4.9x^2)=0

We get rid of parentheses

-4.9x^2-21x+45=0

a = -4.9; b = -21; c = +45;
Δ = b2-4ac
Δ = -212-4·(-4.9)·45
Δ = 1323
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1323}=\sqrt{441*3}=\sqrt{441}*\sqrt{3}=21\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-21\sqrt{3}}{2*-4.9}=\frac{21-21\sqrt{3}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+21\sqrt{3}}{2*-4.9}=\frac{21+21\sqrt{3}}{-9.8}
$