0=20+10t-4.9t^2

Solution for 0=20+10t-4.9t^2 equation:

0=20+10t-4.9t^2

We move all terms to the left:

0-(20+10t-4.9t^2)=0

We add all the numbers together, and all the variables

-(20+10t-4.9t^2)=0

We get rid of parentheses

4.9t^2-10t-20=0

a = 4.9; b = -10; c = -20;
Δ = b2-4ac
Δ = -102-4·4.9·(-20)
Δ = 492
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:

$\sqrt{\Delta}=\sqrt{492}=\sqrt{4*123}=\sqrt{4}*\sqrt{123}=2\sqrt{123}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{123}}{2*4.9}=\frac{10-2\sqrt{123}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{123}}{2*4.9}=\frac{10+2\sqrt{123}}{9.8}
$