How to find the degree of a field extension

I don't quite understand how to find the degree of a field extension. First, what does the notation [R:K] mean exactly?

If I had, for example, to find the degree of $\mathbb Q (\sqrt7)$ over $\mathbb Q$, how would I go about it? And how would it be different from, say, $\mathbb C (\sqrt7)$ over $\mathbb C$?

Would it involve finding the minimal polynomial? In the case of $\mathbb Q (\sqrt7)$, I find the minimal polynomial to be $x^2-7$ which is of degree 2, so would this be the value of $[\mathbb Q (\sqrt7):\mathbb Q]$?

In $\mathbb C $, this polynomial is reducible, so I assume somehow it would need to be reduced to find the minimal polynomial. (I am guessing since $x^2-7=(x^2+1)-8=i-8$ this is the polynomial in $\mathbb C$) Would the degree of this be the degree of$[\mathbb C (\sqrt7):\mathbb C]$?

A field $K$ over a field $F$ is in particular a vector space over $F$, and $[K:F]$ is its dimension. For $F(\alpha)$ it's true that this dimension is the degree of the minimal polynomial of $\alpha$ over $F$, because $K$ then has a basis over $F$ given by $1,\alpha,...,\alpha^{n-1}$. $\mathbb{C}$ is algebraically closed, so all its algebraic extensions are trivial, that is, have degree $1$. But your computation of the minimal polynomial of $\mathbb{C}(\sqrt 7)$ is not correct. It's simply $x-\sqrt 7$, since $\mathbb{C}$ contains a square root of $7$. One more error: $x^2+1$ is not equal to $i$ in $\mathbb{C}[x]$. You're confusing the latter ring with the expression of $\mathbb{C}$ as $\mathbb{R}[t]/(t^2+1)$. In those terms $\mathbb{C}[x]$ is $(\mathbb{R}[t]/(t^2+1))[x]$: the two indeterminates $t$ and $x$ have nothing to do with each other.