Or c^2=1/ε0μ0. https://mathsolver.microsoft.com/es/solve-problem/x%5E%7B2%7D%3D%60frac%7B1%7D%7By%7D

This is a direct result of the wave equation for the electromagnetic fields E and H that can be derived from Maxwell's equations . In dielectric materials you get v=c/√(epsilon_r * µ_r) for the speed where the square root in the denominator is the refractive index. Look up standard textbooks on electromagnetism and optics.

I try to dilucidate it's real meaning/relation. How works the vacuum to include it?

It really depends on your interpretation. From the perspective of classical electrodynamics, one obtains the wave equation to show that there are waves that are able to travel in vacuum with speed c, which are the electromagnetic waves. On the other hand, quantum mechanics and quantum electrodynamics represent the idea that the vacuum itself has the potential to generate massless particles, which are travelling with the speed c. For better understanding, I recommend reading more about Electromagnetic waves, Secondary quantization of electromagnetic fields and Quantum vacuum states.

• It comes from the relation:

c^2 = \frac{1}{\varepsilon_0 \mu_0}

• \varepsilon_0: vacuum permittivity

• \mu_0: vacuum permeability

• When combining Maxwell’s equations, you get a wave equation with speed c, showing light is an electromagnetic wave.