Understand why solvability of Galois group means there is a general formula for polynomial roots. At the end of Galois Theory we have the theorem:
“Over a field of characteristic zero, a polynomial is solvable by radicals if and only if its Galois group is solvable”
I don’t understand how this connects to being able to find a general formula for the roots of a polynomial of degree $n$ in terms of its coefficients.
I see that for some polynomials of degree $5$, its Galois group is not solvable and so the polynomial cannot be solvable by radicals i.e. its roots are not radical expressions and therefore there must be no general formula for the radical roots of a Quintic equation.
However I don’t understand the converse. If a Galois group of a polynomial $f\in K[x]\setminus\{K\}$, is solvable then the polynomial is solvable by radicals i.e. its roots must be radical expressions of elements in the coefficient field $K[x]$. Good. However the polynomial being solvable by radicals doesn’t imply its roots are radical expressions of precisely the coefficients of $f$.
How then do we know Galois group being solvable means there’s a general radical formula for the roots in terms of the polynomials coefficients?
 A: If there were a general formula for solving a degree $n$ polynomial by radicals, then certainly every degree $n$ polynomial would have solvable Galois group: the formula would apply, so the polynomial would be solvable, hence the Galois group would be solvable.
By contrapositive, if there were at least one degree $n$ polynomial that does not have solvable Galois group, then there cannot be a general formula for solving polynomials of degree $n$ by radicals.
One can show, explicitly, that there are polynomials of degree $5$ that have non-solvable Galois group (isomorphic to $A_5$ or to $S_5$). Thus, there can be no general formula for solving quintics by radicals.
In principle, it could be that each polynomial is solvable by radicals but there is no general formula; at least, we would not be able to discard that possibility just from the theorem you quote.
But in fact it turns out this is not the case, which can be checked by looking at the “general degree $n$ equation”, which means working with the polynomial
$$F=(x-x_1)\cdots(x-x_n)$$
in the field $K(x_1,\ldots,x_n)$ of rational functions in $x_1,\ldots,x_n$ with coefficients in $K$. This polynomial has Galois group $S_n$, and so is not solvable by radicals for $n\geq 5$. This means that you cannot have a general formula for degree $n\geq 5$, as that formula would imply the solvability of $F$.
