Solving Op amp Voltage output
Well, we are trying to analyze the following circuit:







simulate this circuit &ndash; Schematic created using CircuitLab

Using KCL, we can write:

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

Using KVL, we can write:

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

Substituting \$(2)\$ into \$(1)\$, gives:

$$
\begin{cases}
\frac{\text{V}_\text{in}-\text{V}_-}{\text{R}_1}+\frac{\text{V}_\text{op}-\text{V}_-}{\text{R}_2}=0\\
\\
\text{I}_\text{op}=\frac{\text{V}_\text{op}-\text{V}_-}{\text{R}_2}+\frac{\text{V}_\text{op}-\text{V}_\text{out}}{\text{R}_3}\\
\\
\frac{\text{V}_\text{op}-\text{V}_\text{out}}{\text{R}_3}=\frac{\text{V}_\text{out}}{\text{R}_4}
\end{cases}\tag3
$$

Using the fact that, in an ideal opamp:

$$\text{V}_-=\text{V}_+=0\tag4$$

Solving for the unknows gives:

$$
\begin{cases}
\text{V}_\text{op}=-\frac{\text{R}_2}{\text{R}_1}\cdot\text{V}_\text{in}\\
\\
\text{I}_\text{op}=-\frac{\text{R}_2+\text{R}_3+\text{R}_4}{\text{R}_1\left(\text{R}_3+\text{R}_4\right)}\cdot\text{V}_\text{in}\\
\\
\text{V}_\text{out}=-\frac{\text{R}_2\text{R}_4}{\text{R}_1\left(\text{R}_3+\text{R}_4\right)}\cdot\text{V}_\text{in}
\end{cases}\tag5
$$



Let's suppose that \$x:=\text{R}_1=\text{R}_2=\text{R}_3=\text{R}_4\$, then equation \$(5)\$ simplifies to:

$$
\begin{cases}
\text{V}_\text{op}=-\text{V}_\text{in}\\
\\
\text{I}_\text{op}=-\frac{3}{2x}\cdot\text{V}_\text{in}\\
\\
\text{V}_\text{out}=-\frac{1}{2}\cdot\text{V}_\text{in}
\end{cases}\tag6
$$