How to find the Thevenin equivalent circuit
Just to label everything:



simulate this circuit &ndash; Schematic created using CircuitLab
(You should have been able to do this with the built-in schematic editor, if you have a PC to work with. Please do so whenever possible. It helps us waste less of our time and helps us communicate with you better.)
Perhaps you've been told that an ideal voltage source has zero resistance and that an ideal current source has infinite resistance. (If not, take it from me.) To work out the equivalent resistance seen via terminals A and B you can simply modify the schematic by removing \$I_1\$ and shorting across \$V_1\$:



simulate this circuit
Note that \$R_1\$ and \$R_2\$ are now in parallel with each other. You can work out that resistance. Once done, that resistance and \$R_3\$ and \$R_4\$ are in a series loop, so just add it up. That's your equivalent resistance.
Finally, to work out the equivalent voltage seen at the terminals (unloaded), return to the first schematic, again. Note that as terminals A and B are unconnected, there's no current flowing from A to B. (It's open.) This also means there's no current in \$R_4\$, so there's no voltage drop across it.



simulate this circuit
I've placed little green arrows to indicate \$0\:\text{A}\$ through the &quot;branch&quot; I want your attention focused on, here. Because no current is flowing between A and B, there can also be no current flowing elsewhere in the same branch. The arrows highlight this fact for you.
Also note that since there is no voltage drop across \$R_4\$, then the voltage at Y must be the same as the voltage at B.
However, \$I_1\$'s current loops through \$R_3\$ causing a voltage drop across it. The polarity is already indicated in the schematic for you. You can compute that voltage difference quite easily and this will tell you the difference between point X and point A.
The only thing left to do is to work out the difference between point Y and point X. But this is determined entirely by \$V_1\$'s voltage applied across the resistor divider formed by \$R_1\$ and \$R_2\$. (You are looking for the voltage drop across \$R_2\$.)
Once you have summed up the voltage drops across \$R_4\$ (no drop at all), \$R_2\$ (computed as part of that voltage divider I just mentioned), and \$R_3\$ (easily worked out using \$I_1\$) then you have the voltage difference between A and B.
Knowing the equivalent voltage (just worked out) and the equivalent resistance (worked out above, earlier) you have your Thevenin voltage and resistance. From that, and by then inserting a \$12\:\Omega\$ resistor across A and B, you can easily work out the current.