getting nodal voltage in a circuit
The circuit is easily redrawn as:







simulate this circuit &ndash; Schematic created using CircuitLab

It's now pretty clear that there is only one unknown node voltage to find. Everything else is definable, as shown.

So, the solution is easy. The currents spilling outward from \$V_x\$ are:

\$\frac{V_x}{R_1=5\Omega} + \frac{V_x}{R_2=2\Omega} + \frac{V_x}{R_3=3\Omega}\$

And the currents spilling inward from other nodes into \$V_x\$ are:

\$\frac{V_1=+25V}{R_1=5\Omega} + 10A + \frac{V_0=0V}{R_2=2\Omega} + \frac{V_2=+7V}{R_3=3\Omega}\$

These must be equal, of course, or else charge would pile up into the \$V_x\$ node. So:

\$\frac{V_x}{R_1=5\Omega} + \frac{V_x}{R_2=2\Omega} + \frac{V_x}{R_3=3\Omega} =
\frac{V_1=+25V}{R_1=5\Omega} + 10A + \frac{V_0=0V}{R_2=2\Omega} + \frac{V_2=+7V}{R_3=3\Omega}\$

\$V_x\cdot\left(\frac{1}{5} + \frac{1}{2} + \frac{1}{3}\right) =
\frac{25V}{5\Omega} + 10A + \frac{0V}{2\Omega} + \frac{7V}{3\Omega}\$

\$V_x =
\frac{5 + 10 + \frac{7}{3}}{\frac{1}{5} + \frac{1}{2} + \frac{1}{3}} = \frac{520}{31}V \approx 16.774V\$

I'm taking you at your word that you worked on it for a while. Hopefully, the above will then make some sense. It's very clear there is only one unknown node voltage. And it's very clear that the currents into and out of that node must be the same.

Personally, I don't like the usual way of instruction using nodal analysis. I prefer the above approach where I "see" currents as spilling away from it and into it. I keep track of things better that way. (It's yet another way to state the superposition principle.)