0=-4.9t^2+22.54t+1.1

Solution for 0=-4.9t^2+22.54t+1.1 equation:

0=-4.9t^2+22.54t+1.1

We move all terms to the left:

0-(-4.9t^2+22.54t+1.1)=0

We add all the numbers together, and all the variables

-(-4.9t^2+22.54t+1.1)=0

We get rid of parentheses

4.9t^2-22.54t-1.1=0

a = 4.9; b = -22.54; c = -1.1;
Δ = b2-4ac
Δ = -22.542-4·4.9·(-1.1)
Δ = 529.6116
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{-(-22.54)-\sqrt{529.6116}}{2*4.9}=\frac{22.54-\sqrt{529.6116}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22.54)+\sqrt{529.6116}}{2*4.9}=\frac{22.54+\sqrt{529.6116}}{9.8}
$