implication of the Abel–Ruffini theorem I am taking a course in abstract algebra, and we proved the following theorem:

I want to prove something more specific. Let's look at polynomials of degree 5 over C.
Someone is claiming he has a magic formula, which receives the coefficients of a polynomial of degree 5, and returns its roots using only basic operations and radicals.
I want to understand, how I can prove this person wrong using the theorem above.
In this case, the theorem talks about the field of rational functions with 5 variables over C. It shows that I can't express $t_1, ..., t_5$ (the roots of f) in terms of $s_1,...,s_5$, in this abstract field.
I understand the proof in this context, but I want to understand how I can use it concretely in order to prove this person wrong.
In the sources that I have seen, they say that Abel–Ruffini theorem implies what I want to prove, but they don't show how. Can someone help me understand how you can show this?
I am adding the proof we saw in the course:

 A: Nice question. I was annoyed by the same thing after learning Galois theory. Before learning it I thought that I will see a proof that there is no formula which works specifically for polynomials over $\mathbb{C}$, but instead of that I only saw a proof that there is no formula which works for polynomials over the field $\mathbb{C}(t_1,...,t_n)$, which is a much bigger field. I was very disappointed for a while, but lucky for me I managed to think of a proof myself.
Suppose there is a formula which works for polynomials of degree $n\geq 5$ over $\mathbb{C}$. The formula contains the field operations, taking roots, the coefficients of a polynomial and some complex constants. (for example the quadratic formula $\frac{-b+\sqrt{b^2-4ac}}{2a}$ uses the constants $2,4,...$). The main thing we should note is that the formula can contain only finitely many constants. Let's call them $z_1,...,z_k$. So our formula is actually a formula over the field $K:=\mathbb{Q}(z_1,...,z_k)$. Note that this field is countable, since it is finitely generated over $\mathbb{Q}$. Since $\mathbb{C}$ is uncountable there must be an element $t_1\in\mathbb{C}$ which is transcendental over $K$. Again, $K(t_1)$ is countable, so there is $t_2\in\mathbb{C}$ which is transcendental over $K(t_1)$. We continue this way, and finally get a field $K(t_1,...,t_n)$ where $t_i$ is transcendental over $K(t_1,...,t_{i-1})$ for all $i$. Now define the polynomial $f=(x-t_1)...(x-t_n)$ and call its symmetric functions $s_1,...,s_n$. (the coefficient of $x^n$ is $1$). Finally, let $L=K(s_1,..,s_n)$, this is a subfield of $K(t_1,...,t_n)$.
Now, what can we say about $L$? I claim that every polynomial of degree $n$ in $L[x]$ is solvable by radicals. Why? Well, take a polynomial $g\in L[x]$ of degree $n$ and put its coefficients in the formula we have. The coefficients of $g$ are obviously in $L$, and remember that the constants in the formula belong to $K\subseteq L$, so they are in $L$ as well! So this shows $g$ is solvable by radicals over $L$.
But now let's go back to the polynomial $f=(x-t_1)(x-t_2)...(x-t_n)\in L[x]$. As we showed above it must be solvable by radicals over $L$, so its Galois group over $L$ is solvable. On the other hand, using the fact that $t_1,...,t_n$ are algebraically independent over $K$ (which means $t_i$ is transcendental over $K(t_1,...,t_{i-1})$ for all $i$) we can conclude that $Gal(K(t_1,...,t_n)/K(s_1,...,s_n))\cong S_n$, this is exactly the same proof as the proof that the Galois group of the general polynomial in the field of rational functions is $S_n$. This means the Galois group of $f$ over $L$ is $S_n$, a contradiction.
The difference is that in my proof $t_1,...,t_n$ are all complex numbers, and not just formal variables like in the field of rational functions. So here $f=(x-t_1)...(x-t_n)$ is a specific polynomial over $\mathbb{C}$ for which the formula we took fails.
