Finding current (I) in a basic circuit
I figured since v1 is giving off a negative current


That doesn't make any sense.  \$v_1\$ is a node voltage and can't be described as 'giving off current' one way or the other.


  I realized there aren't really any nodes, just loops.


There are three nodes.

Honestly, I think you're in over your head for even this simple circuit.  The solution is just a straight forward application of Ohm's law and combining series and parallel resistances.

Hint:  the voltage across the current source is \$v_2 - v_1 = 54V\$.  If you can find the equivalent resistance of the resistors, the current \$i\$ is found by Ohm's law.



Since another answer has been accepted, the solution is, by inspection:

$$i = \frac{v_2 - v_1}{R_{EQ}} = \frac{54}{6||(8+4)}A$$

Note:  you can exchange the position of the current source and 6 ohm resistor to more clearly see how to combine the resistors into the equivalent resistance.



I must comment that node voltage analysis isn't applicable here since the node voltages are given.

If you don't see the equivalent resistance method, then simply use Ohm's law and KCL:







simulate this circuit &ndash; Schematic created using CircuitLab

Using the voltages and resistances given, the resistor currents are easily calculated as shown.

Apply KCL at node 2 to find the current \$i\$:

$$i = 9 + 4.5 = 13.5A$$

which is identical to

$$i = \frac{54}{6||(8+4)} = \frac{54}{4} =  13.5A$$