0=-4.9t^2+0.5t

Solution for 0=-4.9t^2+0.5t equation:

0=-4.9t^2+0.5t

We move all terms to the left:

0-(-4.9t^2+0.5t)=0

We add all the numbers together, and all the variables

-(-4.9t^2+0.5t)=0

We get rid of parentheses

4.9t^2-0.5t=0

a = 4.9; b = -0.5; c = 0;
Δ = b2-4ac
Δ = -0.52-4·4.9·0
Δ = 0.25
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{-(-0.5)-\sqrt{0.25}}{2*4.9}=\frac{0.5-\sqrt{0.25}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-0.5)+\sqrt{0.25}}{2*4.9}=\frac{0.5+\sqrt{0.25}}{9.8}
$