0=-4.9t^2+40t

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

0=-4.9t^2+40t

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

4.9t^2-40t=0

a = 4.9; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·4.9·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*4.9}=\frac{0}{9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*4.9}=\frac{80}{9.8}
=8+1/6.125
$