Ultra simple circuit (2 Resistor, 1 Voltage, 1 Current): Kirchhoffs Laws don't give the solution; Superposition, but how?
Your KVL looks like this:
$$ 
I_{b2}*R_{b2} + I_{b1}*R_{b1}=V_{cc} \\
- I_{b1}+I_{b2}=I_{b}
$$

And if we plug the numbers into WolframAlpha 

http://www.wolframalpha.com/input/?i=15+%3D+X2*82+%2B+X1*15,+X2+-+X1+%3D%3D+40

We get 

\$I_{b1} = -33.66mA\$ and \$I_{B2} = 6.34mA\$ and \$Vb = 15V - 82k\Omega \cdot 6.34mA = - 504.8V \$

Now we can try the superposition 

First I remove the current source, so we are left with this: 







simulate this circuit &ndash; Schematic created using CircuitLab

The voltage divider output voltage is : 

$$V_B' = Vcc*\frac{R_1}{R_2 + R1} = 2.319V$$

Next, we are short-out the voltage source so we are left with this:







simulate this circuit

Therefore \$V_B'' = - R_1||R_2 * I_B = -507.21V\$

And finally 

$$V_B = V_B'+V_B'' = 2.319V + (-507.21V) = -504.8V$$

And we are done.  

As you can see there is something wrong with your equivalent circuit. 

I suspect that the \$I_B\$ (base current) is equal to \$40\mu A\$ (not \$40mA\$) and in this case \$V_b = 1.812V\$ seems reasonable.

Try to show us the original circuit.