0=-4.9t^2+20t+100

Solution for 0=-4.9t^2+20t+100 equation:

0=-4.9t^2+20t+100

We move all terms to the left:

0-(-4.9t^2+20t+100)=0

We add all the numbers together, and all the variables

-(-4.9t^2+20t+100)=0

We get rid of parentheses

4.9t^2-20t-100=0

a = 4.9; b = -20; c = -100;
Δ = b2-4ac
Δ = -202-4·4.9·(-100)
Δ = 2360
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:

$\sqrt{\Delta}=\sqrt{2360}=\sqrt{4*590}=\sqrt{4}*\sqrt{590}=2\sqrt{590}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{590}}{2*4.9}=\frac{20-2\sqrt{590}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{590}}{2*4.9}=\frac{20+2\sqrt{590}}{9.8}
$