I have the following 2 diagrams:
The step response is of magnitude of 3.
I'm a beginner at this so I've done something stupid probably but I have trouble finding answers regarding control engineering on the internet. This is what I tried doing:
I found the poles and the zeros from the root locus. z=-5,+4 p=-6,-10,-3
I think my transfer function is given from this formula but I'm not sure if we have an H(s) in the feedback and it is not stated : $$ T(s)= \frac{KG(s)}{1+KG(s)} $$
From the poles and the zeros my open-loop transfer function G(s) is :
$$ G(s)= \frac{(s+5)(s-4)}{(s+10)(s+6)(s+3)}$$ Doing the calculations I find : $$ T(s)= \frac{Ks^2+Ks-20K}{s^3+(K+19)s^2+(108+K)+180-20K}$$
From the step response(final value is 4) and the final value theorem I find $\frac{-20K}{180-20K}=-4/3=>K=5.14$ I divided 4 by 3 because the diagram above is for a step of magnitude of 3. Now testing the step response of the function I found I get this: 
It's similar to the first response but not identical.
What am I missing here?
You are thinking too complicated. Namely the input to the root locus and step response diagram is the open loop transfer function, so the question is asking you to find this open loop transfer function. The step response diagram tells you the steady state gain, while the root locus diagram the poles and zeros of the transfer function. So the overal transfer function will look like,
$$ G(s) = \frac{K\, (s+5)\, (s-4)\, }{(s+10)\, (s+6)\, (s+3)}, $$
with $3\, G(0) = -4$. Substituting in the equation for $G$ and solving for $K$ yields $K = 12$.