0=-4.9t^2+40t-50

Solution for 0=-4.9t^2+40t-50 equation:

0=-4.9t^2+40t-50

We move all terms to the left:

0-(-4.9t^2+40t-50)=0

We add all the numbers together, and all the variables

-(-4.9t^2+40t-50)=0

We get rid of parentheses

4.9t^2-40t+50=0

a = 4.9; b = -40; c = +50;
Δ = b2-4ac
Δ = -402-4·4.9·50
Δ = 620
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-\sqrt{620}}{2*4.9}=\frac{40-\sqrt{620}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+\sqrt{620}}{2*4.9}=\frac{40+\sqrt{620}}{9.8}
$