# Getting proper Mesh equations

circuit analysis

Here is the circuit I need to solve.

simulate this circuit – Schematic created using CircuitLab

My equations are as follows:

$$Vs_1+8(R_1 +R_3)=0 \\ Vs_2+12I_2-12I_3=0 \\ -28+26I_3+12(I_3-I_2)+8(I_3+I_1)=0$$

Which gives: $$Vs_1=-71.384 \\ Vs_2=-60.923 \\I_3=2.923$$

However these numbers don't make sense. When adding unkowns to equations such as Vs_1 and Vs_2, do you assume they're always positive, or should the have been assigned polarities?

There is no assigned polarity here, so you have choose to one. Generally it is reasonable to choose the voltage drop across a current source to be in the different direction as its current, because normally it tends to deliver power, not use it. For example: \$V_{s1}=V_A-V_D |_{V_D=0}=V_A\$ and \$V_{s2}=V_A-V_B\$.
I assume your first equation was meant to be \$V_{s1}+8(I_1+I_3)=0\$. With my notation (it's a bit more practical, given that \$V_D\$ is ground), it becomes $$1) -V_{s1}+8(I_1+I_3)=0$$
Likewise, your second equation becomes $$2) -V_{s2}+12(I_2-I_3)=0$$
The third and last one has a small mistake, the correct form is: $$3) -28+6I_3+12(I_3-I_2)+8(I_3+I_1)=0$$