0=-4.9t^2+22t+0.5

Solution for 0=-4.9t^2+22t+0.5 equation:

0=-4.9t^2+22t+0.5

We move all terms to the left:

0-(-4.9t^2+22t+0.5)=0

We add all the numbers together, and all the variables

-(-4.9t^2+22t+0.5)=0

We get rid of parentheses

4.9t^2-22t-0.5=0

a = 4.9; b = -22; c = -0.5;
Δ = b2-4ac
Δ = -222-4·4.9·(-0.5)
Δ = 493.8
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{-(-22)-\sqrt{493.8}}{2*4.9}=\frac{22-\sqrt{493.8}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+\sqrt{493.8}}{2*4.9}=\frac{22+\sqrt{493.8}}{9.8}
$