Op-amp output voltage equation
Well, we are trying to analyze the following circuit (assuming an ideal opamp):







simulate this circuit &ndash; Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$
\begin{cases}
\text{I}_1=\text{I}_0+\text{I}_2+\text{I}_4\\
\\
0=\text{I}_2+\text{I}_3\\
\\
\text{I}_3=\text{I}_0+\text{I}_4
\end{cases}\tag1
$$

When we use and apply Ohm's law, we can write the following set of equations:

$$
\begin{cases}
\text{I}_1=\frac{\text{V}_2}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_\text{y}-\text{V}_1}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_3-\text{V}_1}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_\text{x}-\text{V}_3}{\text{R}_4}
\end{cases}\tag2
$$

Substitute \$(2)\$ into \$(1)\$, in order to get:

$$
\begin{cases}
\frac{\text{V}_2}{\text{R}_1}=\text{I}_0+\frac{\text{V}_\text{y}-\text{V}_1}{\text{R}_2}+\frac{\text{V}_\text{x}-\text{V}_3}{\text{R}_4}\\
\\
0=\frac{\text{V}_\text{y}-\text{V}_1}{\text{R}_2}+\frac{\text{V}_3-\text{V}_1}{\text{R}_3}\\
\\
\frac{\text{V}_3-\text{V}_1}{\text{R}_3}=\text{I}_0+\frac{\text{V}_\text{x}-\text{V}_3}{\text{R}_4}
\end{cases}\tag3
$$

Now, when we have an ideal opamp we know that \$\text{V}_\text{k}:=\text{V}_+=\text{V}_-=\text{V}_1=\text{V}_2\$. So we can rewrite equation \$(3)\$ as follows:

$$
\begin{cases}
\frac{\text{V}_\text{k}}{\text{R}_1}=\text{I}_0+\frac{\text{V}_\text{y}-\text{V}_\text{k}}{\text{R}_2}+\frac{\text{V}_\text{x}-\text{V}_3}{\text{R}_4}\\
\\
0=\frac{\text{V}_\text{y}-\text{V}_\text{k}}{\text{R}_2}+\frac{\text{V}_3-\text{V}_\text{k}}{\text{R}_3}\\
\\
\frac{\text{V}_3-\text{V}_\text{k}}{\text{R}_3}=\text{I}_0+\frac{\text{V}_\text{x}-\text{V}_3}{\text{R}_4}
\end{cases}\tag4
$$

Now, for the output voltage we get:

$$\text{V}_3=-\frac{\text{R}_3}{\text{R}_2}\cdot\text{V}_\text{y}\tag{5}$$