0=-4.9t^2+19.6t+104

Solution for 0=-4.9t^2+19.6t+104 equation:

0=-4.9t^2+19.6t+104

We move all terms to the left:

0-(-4.9t^2+19.6t+104)=0

We add all the numbers together, and all the variables

-(-4.9t^2+19.6t+104)=0

We get rid of parentheses

4.9t^2-19.6t-104=0

a = 4.9; b = -19.6; c = -104;
Δ = b2-4ac
Δ = -19.62-4·4.9·(-104)
Δ = 2422.56
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{-(-19.6)-\sqrt{2422.56}}{2*4.9}=\frac{19.6-\sqrt{2422.56}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19.6)+\sqrt{2422.56}}{2*4.9}=\frac{19.6+\sqrt{2422.56}}{9.8}
$