20=40t+4.9t^2

Solution for 20=40t+4.9t^2 equation:

20=40t+4.9t^2

We move all terms to the left:

20-(40t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-40t+20=0

a = -4.9; b = -40; c = +20;
Δ = b2-4ac
Δ = -402-4·(-4.9)·20
Δ = 1992
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1992}=\sqrt{4*498}=\sqrt{4}*\sqrt{498}=2\sqrt{498}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{498}}{2*-4.9}=\frac{40-2\sqrt{498}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{498}}{2*-4.9}=\frac{40+2\sqrt{498}}{-9.8}
$