0=4.9t^2+50t-100

Solution for 0=4.9t^2+50t-100 equation:

0=4.9t^2+50t-100

We move all terms to the left:

0-(4.9t^2+50t-100)=0

We add all the numbers together, and all the variables

-(4.9t^2+50t-100)=0

We get rid of parentheses

-4.9t^2-50t+100=0

a = -4.9; b = -50; c = +100;
Δ = b2-4ac
Δ = -502-4·(-4.9)·100
Δ = 4460
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:

$\sqrt{\Delta}=\sqrt{4460}=\sqrt{4*1115}=\sqrt{4}*\sqrt{1115}=2\sqrt{1115}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{1115}}{2*-4.9}=\frac{50-2\sqrt{1115}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{1115}}{2*-4.9}=\frac{50+2\sqrt{1115}}{-9.8}
$