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I'm trying to find R4 that will make:

$$frac{V_{out}}{V_{in}}=-120$$

The operational amplifier is ideal.

My attempt: In the '-' input there is a virtual ground, so the equivalent resistor above is $$(R2||R3) + R4$$

Using the formula for inverter amplifier:

$$ G = -frac{(R2||R3) + R4}{R1}$$

The answer is approximately $$R4=120MOmega$$

I know from a simulation that the answer is around $$24kOmega$$
Where is my mistake?

^{simulate this circuit – Schematic created using CircuitLab}

Ok, time for idealized opamp rules.

1. No current into opamp inputs

So, all current that flows through R1 also has to flow through R2.

(derivative rule) An opamp in negative feedback will coerce its inputs to take 0V difference.

That means V⁻ will be 0V (GND potential).

From that follows that the right hand side of R2 (where the other resistors attach) is at - (Vin/2), since R1 = 2 R2 (and the same current flows through both).

Since the voltage across R3 is $V_text{node}=-frac12 V_text{in}$, the current through R3 is

$$I_3 = frac{-frac12 V_text{in}}{mathrm R3}text,$$

which leaves us with

begin{align}
I_4 &= I_text{in} - I_3\
&= frac{V_text{in}}{mathrm R1} - frac{-frac12 V_text{in}}{mathrm R3}\
&=V_text{in}left(frac1{mathrm R1}+frac1{2mathrm R3}right)
end{align}
to flow through R4.

That yields an output voltage of

begin{align}
V_text{out} &= V_text{node} + I_4cdot mathrm R_4\
&=V_text{node} +V_text{in}left(frac1{mathrm R1}+frac1{2mathrm R3}right)cdot mathrm R_4\
&= -frac12 V_text{in} +V_text{in}left(frac1{mathrm R1}+frac1{2mathrm R3}right)cdot mathrm R_4\
&= V_text{in}left(frac{mathrm R4}{mathrm R1}+frac{mathrm R4}{2mathrm R3} - frac12right)text.
end{align}

The gain is hence

begin{align}
frac{V_text{out}}{V_text{in}} &= frac{mathrm R4}{mathrm R1}+frac{mathrm R4}{2mathrm R3} - frac12 \
&overset != -120\
-119.5 &= mathrm R4left(frac{1}{mathrm R1}+frac{1}{2mathrm R3}right)
end{align}

I must admit I can't find a non-negative solution for R4, so I must've gone wrong somewhere.

Here is an alternative solution:

Applying the star-triangle-transformation we get thee other resistors.

However, two of them play no role for the closed-loop gain because they are located at the output to ground (pure load) resp. between the inverting input and ground (no influence on closed loop gain).
Hence, there is only one resistor RF left between output and inverting input (one single feedback resistor).

Using the known resistor values it is no problem to find the value of this feedback resistor RF which is a function of the unknown resistor R4.

My result: R4=23.995k-0.09998k=23.895kohms.

One option to determine the gain of this circuit is to use superposition and determine the voltage at the inverting pin which should be 0 V with an idealized op-amp. First, set $V_{out}$ to 0 V and determine $V_{-}$:

$V_{(2a)}=frac{R_4||R_3+R_2}{R_4||R_3+R_2+R_1}V_{in}$

Then, set $V_{in}$ to 0 V and determine again the voltage at $V_{-}$:

Doing the simple maths ok leads to;

$V_{(2b)}=V_{out}frac{R_3}{R_3+R_4}frac{R_1}{R_1+R_2+R_3||R_4}$

Then you say that $V_{-}=V_{(2a)}+V_{(2b)}=0$ and you solve for $V_{out}$ and factor the result. You should find:

$G=-frac{R_2(R_3+R_4)+R_3R_4}{R_1R_3}$

and the value of $R_4$ to have -120 V as an output of this op-amp is given by

$R_4=-frac{R_2R_3-120R_1R_3}{R_2+R_3}=23.895;kOmega$

The below SPICE simulation confirm the value with a perfect op-amp:

Another option would have consisted of using the EET or extra-element theorem which is part of the FACTs but using superposition is already part of the FACTs toolbox.

Given the value of R2 (500 kohm) it will barely be loading the 100 ohm resistor (R3) and so, to obtain a gain of -120, a close approximation (given that R4 will need to be orders of magnitude greater than R3) is to choose a value for R4 that is 120 times R3. But, noting that R2 and R1 have a ratio of 0.5, the real value for R4 is close to 240 times that of R1 i.e. 24 kohm.

Where is my mistake?

I don't recognize your approach.

You could do it more thoroughly by estimating the equivalent Thevenin source voltage and impedance produced by the op-amp output, R4 and R3 of course. Then recognize that the source impedance is in series with R2.