The polynomial $f=3x^5-15x+5$ is irreducible over the rationals, by Eisenstein's criterion. By 19.6 in Gallian, it has no multiple roots. It is not hard to see that $f$ has exactly 3 real roots, so it also has roots $a\pm bi$ for some $a$ and $b$, by problem 73 in chapter 15 of Gallian.
Denote the roots of $f$ by $r_1,\ldots,r_5$, so the splitting field of $f$ is $\Q(r_1,\ldots,r_5)=\Q(\overline r)$. Any $\sigma\in G(\Q(\overline r)/\Q)$ permutes the roots $r_i$, and in fact $\sigma$ is completely determined by its action on the roots. In other words, $G(\Q(\overline r)/\Q)\le S_5$ (there is of course a hidden isomorphism here). Since $[\Q(\overline r)\:\Q]=[\Q(\overline r)\:\Q(r_1)][\Q(r_1)\:\Q]= [\Q(\overline r)\:\Q(r_1)]\cdot 5$, $[\Q(\overline r)\:\Q]= |G(\Q(\overline r)/\Q)|$ is divisible by 5.
Lemma 2.3.1 If $p$ is prime and $p$ divides $|G|=n$, then $G$ has an element of order $p$.
Proof. What elements of $G$ are solutions to $x^p=1$? Certainly $x=1$ is a root. If $y\not=1$ is a solution, then $y$ has order $p$, so it suffices to show that there are at least 2 solutions.
Let $S=\{(a_1,\ldots,a_p)|\prod a_i=1\}$. For any choice of $a_1,\ldots,a_{p-1}$, there is a unique $a_p$ such that $(a_1,\ldots,a_p)$ is in $S$, so $|S|=n^{p-1}$. If $\overline a$ and $\overline b$ are in $S$, say $\overline a\equiv \overline b$ if each may be obtained by rotating the other—for example, $(a_1,\ldots,a_p)\equiv (a_3,\ldots,a_p,a_1,a_2)$. If $a_i=a_j$ for all $i$ and $j$, then $(a_1,\ldots,a_p)$ forms an entire equivalence class. Otherwise, since $p$ is prime, there must be exactly $p$ members of each equivalence class. Each solution $c$ of $x^p=1$ corresponds to an equivalence class $\{(c,c,c,\ldots,c)\}$, and vice versa, so the number of solutions is equal to the number of one element equivalence classes, say $r$. Let $s$ be the number of $p$-element equivalence classes. Then $n^{p-1}=r+sp$. By hypothesis $p|n$, so $p|r$. Since $r\ge 1$, this means $r\ge 2$ as desired. $\qed$
Returning to $G(\Q(\overline r)/\Q)$, we now see that this group contains an element of order 5, which must be a 5-cycle. Let $\sigma\colon \C\to \C$ be defined by $\sigma(x+yi)=x-yi$; $\sigma$ is an automorphism (example 2, chapter 15). Since $\sigma$ permutes the roots of $f$, its restriction to $\Q(\overline r)$ is in $G(\Q(\overline r)/\Q)$, and clearly $\sigma$ has order 2. Thus, $G(\Q(\overline r)/\Q)$ contains both a 2-cycle and a 5-cycle, say $(b_1,b_2,b_3,b_4,b_5)$ and $(c_1,c_2)$. It is not hard to see that we may assume that the 2-cycle is $(b_1,b_2$. It is then not hard to show that $G(\Q(\overline r)/\Q)$ must contain a 5-cycle, a 4-cycle and a 3-cycle, meaning $|G(\Q(\overline r)/\Q)|\ge 60$, so the order is either 60 or 120. Finally, since there is only one subgroup of order 60, namely $A_5$ (exercise 24.21 in Gallian), and since $A_5$ contains no 2-cycle, $G(\Q(\overline r)/\Q)$ must be $S_5$. Since $S_5$ is not solvable, $f$ is not solvable by radicals. Since this particular quintic is not solvable by radicals, it is clear also that there can be no "formula'' for the roots of a quintic.
In fact, not even one of the roots may be written as an expression in radicals, since if it could, we could divide out the corresponding linear factor and then solve the resulting quartic completely, giving a complete solution of the quintic in radicals. This argument works only for fifth degree polynomials; here is a more general way to talk about individual roots.
Theorem 2.3.2 Suppose $f\in F[x]$ has roots $r_1,\ldots,r_n$, and that for all $i$, $F(r_1)\cong F(r_i)$, by an isomorphism $\sigma_i$ that fixes $F$. If there is a radical extension of $F$ containing $F(r_1)$ then there is a radical extension of $F$ in which $f$ splits, that is, $f$ is solvable by radicals.
Proof. Let $F_i=F_{i-1}(a_i)$, $i=1,\ldots,s$, be the tower of fields forming the radical extension containing $F(r_1)$. Let $g_i$ be the minimal polynomial of $a_i$ over $F$, and let $g=f\prod g_i$. Let $N$ be the splitting field of $g$ over $F$. By theorem 19.4 in Gallian, the isomorphism $\sigma_i$ from $F(r_1)$ to $F(r_i)$ can be extended to an automorphism on $N$. Then $\sigma_i(F_s)$ is a radical extension containing $F(r_i)$, formed using the elements $\sigma_i(a_j)$. As in lemma 2.1.4, each of the other roots of $g$ is also contained in a copy of $F_s$, and we may put all of these radical extensions together into a single tower of fields culminating in $N$. Thus $N$ is a radical extension of $F$ that contains the splitting field of $f$, as desired. $\qed$
So as long as the roots of a polynomial "look alike,'' if one of them can be written as an expression in radicals, all of them can and the corresponding Galois group is solvable. In particular, if the Galois group is $S_n$, $n\ge 5$, then the roots are individually not expressible by radicals.