I'm reading Ian Stewart's book "Galois Theory" and even if I can understand definition formally, I have problem understanding it in the sense of application.
Definition: An extension $L:K$ is radical if $L = K(\alpha_1, \alpha_2, \ldots, \alpha_m)$, where for each $i = 1, \ldots, m$ there exists an integer $n(i)$ such that $\alpha_i^{n(i)} \in K(\alpha_1, \ldots, \alpha_{i - 1})$. The elements $\alpha_i$ are said to form a radical sequence for $L:K$.
Definition: Let $f$ be a polynomial over a field $K$ of characteristic $0$, and let $\Sigma$ be a splitting field for $f$ over $K$. We say that $f$ is soluble by radicals if there exists a field $M$ containing $\Sigma$ such that $M:K$ is a radical extension.
And now comes my missunderstanding: let $f$ be a polynomial over $\mathbb{C}$, its splitting field should be $\mathbb{C}$, right? What is its "$M$" then? We could try this with example, let $f(x) = x^2 + 1$, it is a polynomial over $\mathbb{Q}$ and it splits over $\mathbb{Q}[i]$, right? Then what is its $M$? Or consider a polynomial $f(x) = x^2 - 1$ over $\mathbb{Q}$ and it also splits over $\mathbb{Q}$, so what is its $M$?
You are correct that if $f(x)\in\mathbb{C}[x]$, then the roots of $f$ are all in $\mathbb{C}$, so they are in a radical extension of $\mathbb{C}$. The insolvability of the quintic generally refers to quintic polynomials in $\mathbb{Q}[x]$. In this case, the splitting field is certainly contained in $\mathbb{C}$, but not all elements of $\mathbb{C}$ can be expressed in terms of radicals (indeed, the theorem that there is no "quintic formula" tells us not all algebraic numbers over $\mathbb{Q}$ can be expressed in terms of radicals). The point is the roots of the polynomial are in some algebraic extension, but it need not be a radical extension; that is to say, the roots of polynomial with rational coefficients cannot necessarily be expressed in terms of radicals of rationals (as opposed to reals or complex numbers).
For your examples, $f(x)=x^2+1$, then $M=\Sigma=\mathbb{Q}(i)$. As for $f(x)=x^2-1$, this splits over $\mathbb{Q}$, so $M=\Sigma=\mathbb{Q}$. I am not aware of any examples where $M\neq\Sigma$ off the top of my head, but perhaps someone else can chime in with one.