I'm trying to know the fr equation for the following op amp configuration:
[![enter image description here](https://i.stack.imgur.com/GdW0v.jpg)](https://i.stack.imgur.com/GdW0v.jpg)
I have some capacitors values for specific fr frequencies:

fr = 34Hz -> C = 33nF

fr = 50Hz -> C = 22nF

fr = 160Hz -> C = 6.8nF

fr = 330Hz -> C = 3.3nF

fr = 1.1kHz -> C = 1nF

fr = 2.3kHz -> C = 470pF

fr = 6kHz -> C = 180pF

fr = 16kHz -> C = 33nF

Capacitors are not exact values, they are commercial. I know these values are correct because I saw the frequency response in an oscilloscope.

I was searching on web and found this:
[![enter image description here](https://i.stack.imgur.com/3gR3b.png)](https://i.stack.imgur.com/3gR3b.png)

This is the page where I found it: <https://www.electronics-tutorials.ws/filter/filter_7.html>
The circuits are very similar. The difference is R9 resistor which add an offset to the input signal. This offset is to use just one power source instead of two.
The page give me an equation to calculate the fr frequency.
[![enter image description here](https://i.stack.imgur.com/qAByu.png)](https://i.stack.imgur.com/qAByu.png)

But when I replace the values in the equation, it doesn't give a close values as the practical capacitance.

fr = 15.25Hz -> C = 33nF

fr = 22.87Hz -> C = 22nF

fr = 74.01Hz -> C = 6.8nF

If I multiply the equation by 2, it give me a closer value.

fr = 30.5Hz -> C = 33nF

fr = 45.74Hz -> C = 22nF

fr = 148.02Hz -> C = 6.8nF

If it is the solution, I would like to understand why.

If someone could explain me the first op amp configuration, I would be very grateful.