20=-4.9t^2+25t+0

Solution for 20=-4.9t^2+25t+0 equation:

20=-4.9t^2+25t+0

We move all terms to the left:

20-(-4.9t^2+25t+0)=0

We get rid of parentheses

4.9t^2-25t-0+20=0

We add all the numbers together, and all the variables

4.9t^2-25t+20=0

a = 4.9; b = -25; c = +20;
Δ = b2-4ac
Δ = -252-4·4.9·20
Δ = 233
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{-(-25)-\sqrt{233}}{2*4.9}=\frac{25-\sqrt{233}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{233}}{2*4.9}=\frac{25+\sqrt{233}}{9.8}
$