20=4.905t^2

Solution for 20=4.905t^2 equation:

20=4.905t^2

We move all terms to the left:

20-(4.905t^2)=0

We get rid of parentheses

-4.905t^2+20=0

a = -4.905; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-4.905)·20
Δ = 392.4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{392.4}}{2*-4.905}=\frac{0-\sqrt{392.4}}{-9.81}
=-\frac{\sqrt{}}{-9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{392.4}}{2*-4.905}=\frac{0+\sqrt{392.4}}{-9.81}
=\frac{\sqrt{}}{-9.81}
$