0=-4.9t^2+40t+100

Solution for 0=-4.9t^2+40t+100 equation:

0=-4.9t^2+40t+100

We move all terms to the left:

0-(-4.9t^2+40t+100)=0

We add all the numbers together, and all the variables

-(-4.9t^2+40t+100)=0

We get rid of parentheses

4.9t^2-40t-100=0

a = 4.9; b = -40; c = -100;
Δ = b2-4ac
Δ = -402-4·4.9·(-100)
Δ = 3560
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:

$\sqrt{\Delta}=\sqrt{3560}=\sqrt{4*890}=\sqrt{4}*\sqrt{890}=2\sqrt{890}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{890}}{2*4.9}=\frac{40-2\sqrt{890}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{890}}{2*4.9}=\frac{40+2\sqrt{890}}{9.8}
$