Nodal Analysis using voltage sources
Well, I am trying to analyze the following circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\text{I}_x+\text{I}_2\\
\\
\text{I}_x=\text{I}_3+\text{I}_4\\
\\
\text{I}_1=\text{I}_2+\text{I}_5\\
\\
\text{I}_5=\text{I}_3+\text{I}_4
\end{cases}\tag1
$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_2}{\text{R}_4}
\end{cases}\tag2
$$
We also know that \$\text{V}_x=\text{V}_2-\text{V}_1\$.
Now, because you've to do the math to solve this. I will present a Mathematica code that will solve this problem:
In[1]:=FullSimplify[
 Solve[{I1 == Ix + I2, Ix == I3 + I4, I1 == I2 + I5, I5 == I3 + I4, 
   I1 == (Vi - V1)/R1, I2 == (V1)/R2, I3 == (V2)/R3, I4 == (V2)/R4, 
   Vx == V2 - V1}, {Ix, I1, I2, I3, I4, I5, V1, V2}]]

Out[1]={{Ix -&gt; ((R3 + R4) (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I1 -&gt; (R3 R4 Vi + R2 (R3 + R4) (Vi + Vx))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I2 -&gt; (R3 R4 Vi - R1 (R3 + R4) Vx)/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I3 -&gt; (R4 (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I4 -&gt; (R3 (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  I5 -&gt; ((R3 + R4) (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  V1 -&gt; (R2 R3 R4 Vi - R1 R2 (R3 + R4) Vx)/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4), 
  V2 -&gt; (R3 R4 (R1 Vx + R2 (Vi + Vx)))/(
   R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4)}}

Now, using your values we get:
In[2]:=R1 = 4;
R2 = 3;
R3 = 2;
R4 = 6;
Vi = 14;
Vx = 6;
FullSimplify[
 Solve[{I1 == Ix + I2, Ix == I3 + I4, I1 == I2 + I5, I5 == I3 + I4, 
   I1 == (Vi - V1)/R1, I2 == (V1)/R2, I3 == (V2)/R3, I4 == (V2)/R4, 
   Vx == V2 - V1}, {Ix, I1, I2, I3, I4, I5, V1, V2}]]

Out[2]={{Ix -&gt; 56/15, I1 -&gt; 18/5, I2 -&gt; -(2/15), I3 -&gt; 14/5, I4 -&gt; 14/15, 
  I5 -&gt; 56/15, V1 -&gt; -(2/5), V2 -&gt; 28/5}}

So, the answers are \$\text{v}=\text{V}_1=-\frac{2}{5}=-0.4\space\text{V}\$ and \$\text{i}=\text{I}_3=\frac{14}{5}=2.8\space\text{A}\$.