0=98+49t-4.9t^2

Solution for 0=98+49t-4.9t^2 equation:

0=98+49t-4.9t^2

We move all terms to the left:

0-(98+49t-4.9t^2)=0

We add all the numbers together, and all the variables

-(98+49t-4.9t^2)=0

We get rid of parentheses

4.9t^2-49t-98=0

a = 4.9; b = -49; c = -98;
Δ = b2-4ac
Δ = -492-4·4.9·(-98)
Δ = 4321.8
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{-(-49)-\sqrt{4321.8}}{2*4.9}=\frac{49-\sqrt{4321.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-49)+\sqrt{4321.8}}{2*4.9}=\frac{49+\sqrt{4321.8}}{9.8}
$