Nodes A and B forms a supernode with the reference node due to the 12 V source and the 25 V source (shown with a dotted outline):

V_A = 12

V_A - V_B = 25

12 - V_B = 25

V_B = - 13

KCL at node C:

Simplify and collect terms

12 V_A - 12 V_C + 6 V_B - 6 V_C + 3 V_D - 3 V_C - 4 V_C = 0

12 (12) + 6 (-13) - 25 V_C + 3 V_D = 0

- 25 V_C + 3 V_D = - 66 (Eq. 1)

 

KCL at node D:

Simplify and collect terms (LCD = 60)

12 (-13) - 12 V_D + 15 V_C - 15 V_D - 10 V_D = 0

15 V_C - 37 V_D = 156 (Eq. 2)

Solving Equations 1, and 2 yields:

V_A = 12 V

V_B = -13 V

V_C = 2.24 V

V_D = - 3.31 V