Here is an application of that formula to a problem similar to yours above:
There is no extension of $\Bbb{R}$ of degree 4.
Suppose we have an extension $F$ of the reals of degree 4. Then such an extension $F$ is $\Bbb{R}[a_1,\ldots a_n]$ where the $a_i \in F$ (why?). Now since $F$ is an extension of $\mathbb{R}$ of degree 4 we are adjoining at most two elements of $F$ to $\Bbb{R}$. This is because if say we adjoin three elements $a,b,c$ to $\Bbb{R}$, without loss of generality we have
$[\Bbb{R}(a,b,c) : \Bbb{R}] = [\Bbb{R}(a,b,c) : \Bbb{R}(a,b) ] [\Bbb{R}(a,b), \Bbb{R}(a) ] [\Bbb{R}(a):\Bbb{R}] $
or that we have 3 positive integers multiplying together to give 4. Therefore one of those extensions must have been the trivial extension. Similarly adjoining 4 or more elements does not create any bigger field than adjoining two elements. We now show that assuming $F$ is an extension of $\Bbb{R}$ by adjoining one or two elements of $F$ leads to a contradiction. Suppose $F = \Bbb{R}(a)$. Then this means that $a \in F$ is the root of a real monic irreducible polynomial of degree 4. However this is not possible because every real polynomial of degree 4 is reducible over $\Bbb{R}$ (exercise; use the fundamental theorem of algebra). So the only possibility left is that $F$ is obtained from $\Bbb{R}$ by adjoining two elements, which we call $\alpha $ and $\beta$. Write $[\Bbb{R}(\alpha,\beta) : \Bbb{R}] = [\Bbb{R}(\alpha,\beta):\Bbb{R}(\alpha)][\Bbb{R}(\alpha):\Bbb{R}].$
For the right hand side to equal 4 this means that each factor in the right hand side is 2. We cannot have one of them being $1$ because then the other is 4 contradicting the fact that all real polynomials are reducible over $\Bbb{R}$. Now if $[\Bbb{R}(\alpha):\Bbb{R}]$ is 2 then this means that $\alpha$ is a root of a real polynomial of degree 2, so that $\alpha$ is contained in the complex numbers, hence $\Bbb{R}(\alpha) \subset \Bbb{C}$. But then we know that there are no proper fields of finite degree as an extension of $\Bbb{R}$ between $\Bbb{C}$ and $\Bbb{R}$. This forces $\Bbb{R}(\alpha) = \Bbb{C}$.
It follows that the extension $\Bbb{R}(\alpha,\beta)/\Bbb{R}(\alpha)$ is an extension of $\Bbb{C}$ of degree two. However this means that $\beta$ is the root of an irreducible polynomial of degree 2 with complex coefficients. However since $\Bbb{C}$ is algebraically closed the only irreducible polynomials in $\Bbb{C}$ are those of degree 1. This is a contradiction so that the right hand side of equation $(1)$ can never equal 4. It follows that there is no field extension of $\Bbb{R}$ of degree 4.