0=20+10t-4.905t^2

Solution for 0=20+10t-4.905t^2 equation:

0=20+10t-4.905t^2

We move all terms to the left:

0-(20+10t-4.905t^2)=0

We add all the numbers together, and all the variables

-(20+10t-4.905t^2)=0

We get rid of parentheses

4.905t^2-10t-20=0

a = 4.905; b = -10; c = -20;
Δ = b2-4ac
Δ = -102-4·4.905·(-20)
Δ = 492.4
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{-(-10)-\sqrt{492.4}}{2*4.905}=\frac{10-\sqrt{492.4}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{492.4}}{2*4.905}=\frac{10+\sqrt{492.4}}{9.81}
$