Well, we have the following circuit (and we assume an ideal model of an OPAMP):

<!-- Begin schematic: In order to preserve an editable schematic, please
     don't edit this section directly.
     Click the "edit" link below the image in the preview instead. -->

![schematic](https://i.sstatic.net/YmTWG.png)

<!-- End schematic -->

Using KCL, we can write:

$$
\begin{cases}
I_4=I_-+I_2\\
\\
I_x=I_++I_3\\
\\
I_\text{o}=I_4+I_5
\end{cases}\tag1
$$

Using KVL, we can write:

$$
\begin{cases}
I_x=\frac{V_x-V_+}{R_1}\\
\\
I_2=\frac{V_--0}{R_2}\\
\\
I_3=\frac{V_+-0}{R_3}\\
\\
I_4=\frac{V_\text{o}-V_-}{R_4}\\
\\
I_5=\frac{V_\text{o}-0}{R_5}\\
\\
\end{cases}\tag2
$$

***Notice***: in the ideal OPAMP circuit we assume that \$I_+=I_-=0\$. If (and only if) the opamp is used in a circuit with negative feedback then we can assume that \$V_+=V_-\$.

Now, the gain is defined by:

$$G:=\frac{V_\text{o}}{V_x}\tag3$$

We can find an expression for the output voltage \$V_\text{o}\$, by solving the systems of equations:

$$V_\text{o}=\frac{V_xR_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}\tag4$$

So, we get:

$$G=\frac{1}{V_x}\times\frac{V_xR_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}=\frac{R_3\left(R_2+R_4\right)}{R_2\left(R_1+R_3\right)}\tag5$$

>In your case we get:

>$$G=\frac{68000\times\left(30000+63000\right)}{30000\times\left(12000+68000\right)}=\frac{527}{200}=2.635\tag6$$

____

Solving it, in general, gives (notice that \$V_+=V_-=V_\text{p}\$):

[![enter image description here][1]][1]

In your case (using your values):

[![enter image description here][2]][2]

>I checked my solution using LTspice and I got it right.


  [1]: https://i.sstatic.net/5bHt7.png
  [2]: https://i.sstatic.net/3KdmJ.png