Can't find inverse Z transform

[Addez123]

Homework Statement

I got the laplace transfer function H(s) = 1/(s + 2) and I'm suppose to find the inverse Z transform by first converting to H(z) by s = Ts/2*(z-1)/(z+1)
Then do inverse Z-transform using the "displacement rule" - Never heard of.

Homework Equations

H(s) = 1/(s + 2)
s = Ts/2*(z-1)/(z+1)

3. The Attempt at a Solution
I can't get any serious answer, and the partial z-inverse that I manage to find it's incredibly complicated (see image).

What is this 'displacement rule' and how do I use it here?

 

[FactChecker]

I don't have reference 17 (James G. Advanced modern engineering mathematics. Reading: Addison-Wesley; 1993.) that they refer to, but it looks like this paper ( https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5240402/ ) uses it to associate the scaled/shift inside δ(γk-λ_tot(t)) of equation 13 with (1/γ)z^{-λ_tot(t)/γ} in equation 15.
Conversions like that are fairly common in transformations involving the delta function.

 

[Addez123]

The solution was to, after replacing 's' with the Z components, then put all under one fraction sign (no plus in nominator). Then multiply with Z^-1 in bottom and top until you got only Z^- terms. This = Y(x)/X(x) which can then be translated back to y[n] and y[n-1] = Z^-1*Y(x)