* Assume that the currents are moving away from the nodes, 

Apply Kirchhof's current Law at Node A

$\dfrac{V_A}{2}+\dfrac{V_A-V_B}{5}=1$

$\Bigg(\dfrac{1}{2}+\dfrac{1}{5}\Bigg)V_A-\dfrac{1}{5}V_B=1\\ 0.7\ v_A-0.2\ V_B=1\ \ \ \ ................(1)$

Apply Kirchhof's current Law at Node B

$\dfrac{V_B-V_A}{5}+\dfrac{V_B}{20}+\dfrac{V_B-V_C}{4}=0\\ -\dfrac{1}{5}V_A+\Bigg(\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{4}\Bigg)V_B-\dfrac{1}{4}V_c=0\\ -0.2\ V_A+0.5\ V_B-0.25\ V_C=0 \ \ \ \ \ \ \ .......................(2)$

Node C is directly connected to a voltage source of 5 V. Hence we can write voltage equation at Node C. 

$V_C=-5.....................(3)$

Solving equations (1), (2) and (3) 

$V_A=0.81\ V\\ V_B=-2.18\ V\\ V_C=-5\ V$