Sallen Key Butterworth Response
Your sallen key filter has a gain of 1 hence it posseses this transfer function: -

\$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}\$

So, if \$\omega_n\$ (the natural resonant frequency) is normalized to 1 you get: -

\$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{s^2+2\zeta s+1}\$ where \$2\zeta = \dfrac{1}{Q}\$

If your Q = 0.707, the inverse is 1.414 (as seen in your polynomial). 

It doesn't matter what value \$\omega_n\$ actually is; for a butterworth response Q = \$\dfrac{1}{\sqrt2}\$

Q is always the transfer function gain at \$\omega_n\$ and, for a butterworth response the gain at \$\omega_n\$ is always -3dB or 0.7071. This produces zero peaking in the pass-band i.e. is maximally flat in the pass-band: -



Picture source

The natural resonant frequency (\$\omega_n\$) in your circuit is \$\dfrac{1}{\sqrt{R_1R_2C_1C_2}}\$. 

I added this point just in case you were wondering about it. 

The full TF of your circuit is: -

 

From here