Worth upgrading at all or not?

Hey...
I basically know the answer, but I want to hear your thoughts on it.
I have Asrock FM2A88X-Extreme6+ with Athlon 760k.

http://www.asrock.com/mb/AMD/FM2A88X%20Extreme6+/
I can get Asrock Fatal1ty 970 Performance with FX6300 for about 25-30$.

http://www.asrock.com/mb/AMD/Fatal1ty%20970%20Performance/
Is it worth upgrading at all, if my goal is to go Zen when it comes out?
Currently I am only gaming on it, but I have some plans I am trying to put in motion, that 6300 may help, but they may not go through until Zen anyway..
Will 6300 help for anything, considering what I am doing with it..

If the total cost is $30 sure why not

if the cost is seriously $30 just do it. the 6core will help slightly but not much since games use 4cores the other 2 cores can handle background stuff such as google chrome being open or something. but again for $30 just do it would be silly not to.

I like how you said chrome, like there are no other options..
Well, the price may vary, but still... It should not surpass 50-60$...
My question is more is it worth for less than an year of usage? Because I can get better psu or gpu upgrade or something and still prepare myself for Zen...

I don't see why not, you can always resell it and the parts you're replacing.

i dont think that it will be a big step up.
depends on your current gpu mainaly.

both chips are basicly build on the same architecture

That's exactly the thing. For what I'm currently using it, it works just fine. It will help with some future plans, but...
It is basically I really want this VS this is a pure waste of 30-40-50$...

OK, i am confused...
The Asrock motherboard have M.2 slot. It is directly wired into the 5-th PCI-E x4 slot... Now:
Isn't M.2 standard suppose to bl;ow SATA out of the water when it comes to speed, or it's like mSATA, just another form factor for the same speed? If that is true, why the hell does it takes the bandwidth of an entire x4 PCI-e, when it can handle so much more data? Or am i completely missing something.