Starfall's answer is brilliant, as always! In addition to Starfall's answer, can I share a concrete example that I personally found helpful when I was learning this subject?
Consider the polynomial $X^n - 2$ over $\mathbb Q$. The splitting field is $\mathbb Q(\zeta_n, \sqrt[n]{2})$.
There is a sequence of normal field extensions:
$$ \mathbb Q \subset \mathbb Q (\zeta_n) \subset \mathbb Q(\zeta_n , \sqrt[n]{2}).$$
Correspondingly, there is a sequence of group inclusions:
$$ 1 \ \lhd \ {\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q(\zeta_n)) \ \lhd \ {\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q).$$
Now, notice that:
${\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q(\zeta_n))$ is abelian, because it is a subgroup of the additive cyclic group $\mathbb Z_n$. Indeed, all automorphisms in ${\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q(\zeta_n))$ send $\sqrt[n]{2} \mapsto \zeta_n^a \sqrt[n]{2}$ for some $a \in \mathbb Z_n$.
The quotient ${\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q) / {\rm Gal}(\mathbb Q (\zeta_n, \sqrt[n]{2}):\mathbb Q(\zeta_n))$ is isomorphic to ${\rm Gal}(\mathbb Q (\zeta_n) : \mathbb Q)$, by the Galois correspondence. This is also abelian, because it is isomorphic to $\mathbb Z_n^\times$, the multiplicative group of units modulo $n$: the automorphisms in ${\rm Gal}(\mathbb Q (\zeta_n) : \mathbb Q)$ are of the form $\zeta_n \mapsto \zeta_n^b$ for $b \in \mathbb Z_n^\times$.
Thus we have explicitly verified that ${\rm Gal}(\mathbb Q(\zeta_n, \sqrt[n]{2} : \mathbb Q)$ is a solvable group!
For more complicated examples, we may need to consider a longer sequence of field extensions. (We build our sequence by successively adjoining an $n$th root of an element of a field previously constructed, so the length of the sequnece will depend on how many times we need to adjoin $n$th roots until we've constructed the splitting field of the polynomial.) We may even wish to adjoin all $n$th roots of unity by hand, at the very beginning. But whatever example we come up with, the idea will always be fundamentally the same as in the example we just worked through: in the corresponding sequence of group inclusions, all quotients will be subgroups of the additive group $\mathbb Z_n$, except for the final one which will be isomorphic to ${\rm Gal}(\mathbb Q(\zeta_n):\mathbb Q)$, the multiplicative group of units modulo $n$.
Of course, this answer only addresses the "solvable by radicals" $\implies$ "solvable Galois group" implication. As Starfall explained, the reverse implication works too - and that properly addresses your original question. Nonetheless, I do find the forward implication more intuitive!