How is the formula of the cutoff frequency of a 2nd order Butterworth low-pass filter derived? According to articles and notes, the cutoff frequency of a 2nd order Butterworth lowpass filter:


is expressed by the formula:
$\omega _c = {1\over\sqrt{R_1R_2C_1C_2}}$

and since $\omega _c = 2 \pi f_c$, alternatively

$f_c={1\over2\pi\sqrt{R_1R_2C_1C_2}}$

Where does this formula come from? I'm asked to write the equations and formulas which eventually lead to the formula of the cutoff frequency, but I've been very confused and tutorials and notes don't help at all. Thanks in advance.
First you need to find (calculate or look up) the transfer function of that circuit. It has the form

$$H(s)=\frac{a}{s^2+bs+c}\tag{1}$$

where the constants $a$, $b$, and $c$ depend on the values of the resistors and capacitors.

The $3$ dB cut-off frequency $\omega_c$ can be found by solving

$$\big|H(j\omega_c)\big|^2=\frac12\big|H(0)\big|^2=\frac12\left(\frac{a}{c}\right)^2\tag{2}$$

(because $H(0)=\frac{a}{c}$).

Solving $(2)$ involves solving a quadratic equation in $\omega_c^2$.