Nodal analysis how to determine high and low potential for current
Easiest way is to ignore the arrows.  Deal with arrows after you have done calculations.

Assume all currents (-ve) are heading out.  Aside from the given current source.

Node D (given)
$$V_D = 0V$$

Node A (given)
$$V_A = 12V$$

Node B
$$+I_4 - I_3 - I_1 = 0$$
$$+I_4 - \frac {V_B - V_C}{R_3} - \frac {V_B - V_A}{R_1} = 0$$
$$+6mA - \frac {V_B - V_C}{910\Omega} - \frac {V_B - +12V}{4.2k\Omega} = 0$$
$$+6mA - \frac {1}{910\Omega}V_B + \frac {1}{910\Omega}V_C - \frac {1}{4.2k\Omega}V_B + \frac {12V}{4.2k\Omega} = 0$$
$$-0.001337 V_B + 0.001099 V_C = -8.857mA\ \ \ \ (1)$$

Node C
$$-I_3 - I_2 - I_5 = 0$$
$$-\frac {V_C - V_B}{910\Omega} - \frac {V_C - 12V}{1.2k\Omega} - \frac {V_C}{3.8k\Omega} = 0$$
$$\frac {1}{910\Omega}V_B - \left ( \frac{1}{910\Omega} + \frac {1}{1.2k\Omega} + \frac {1}{3.8k\Omega} \right )V_C = -10mA$$
$$0.001099 V_B - 0.002195 V_C = -10mA\ \ \ \ (2)$$

Two equations, two unknowns.

$$V_B = 17.62V\ \ \ \ V_C = 13.37V$$