20=12x+4.9x^2

Solution for 20=12x+4.9x^2 equation:

20=12x+4.9x^2

We move all terms to the left:

20-(12x+4.9x^2)=0

We get rid of parentheses

-4.9x^2-12x+20=0

a = -4.9; b = -12; c = +20;
Δ = b2-4ac
Δ = -122-4·(-4.9)·20
Δ = 536
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:

$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{134}}{2*-4.9}=\frac{12-2\sqrt{134}}{-9.8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{134}}{2*-4.9}=\frac{12+2\sqrt{134}}{-9.8}
$