0=100+2t-4.9t^2

Solution for 0=100+2t-4.9t^2 equation:

0=100+2t-4.9t^2

We move all terms to the left:

0-(100+2t-4.9t^2)=0

We add all the numbers together, and all the variables

-(100+2t-4.9t^2)=0

We get rid of parentheses

4.9t^2-2t-100=0

a = 4.9; b = -2; c = -100;
Δ = b2-4ac
Δ = -22-4·4.9·(-100)
Δ = 1964
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:

$\sqrt{\Delta}=\sqrt{1964}=\sqrt{4*491}=\sqrt{4}*\sqrt{491}=2\sqrt{491}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{491}}{2*4.9}=\frac{2-2\sqrt{491}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{491}}{2*4.9}=\frac{2+2\sqrt{491}}{9.8}
$