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

Solution for 0=20(t)-4.9(t^2) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

4.9t^2-20t=0

a = 4.9; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·4.9·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*4.9}=\frac{40}{9.8}
=4+0.8/9.8
$