Roadmap for work below: It will suffice to show that the polynomial has an unsolvable Galois group, namely $S_5$. To do this, we will show that the Galois group, when viewed as a permutation group, has a $5$-cycle and a $2$-cycle; these serve as a generating set for $S_5$.
Let $K$ be the splitting field of $f$. Going forward, it will be helpful to think of the Galois group $\text{Gal}(K/\mathbb{Q}) \subseteq S_5$ as a permutation group acting on the five roots of $f$.
Now, since $f$ is an irreducible quintic, we can adjoin one of its real roots to $\mathbb{Q}$ to yield a degree-$5$ extension of $\mathbb{Q}$, giving the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\alpha) \subset K$. Applying the multiplicativity formula, $|\text{Gal}(K/\mathbb{Q})| = [K:\mathbb{Q}] = 5\cdot[K:\mathbb{Q}(\alpha)]$, so we see that $5$ divides $|\text{Gal}(K/\mathbb{Q})|$. Therefore, by Cauchy's theorem, there must exist an element in $\text{Gal}(K/\mathbb{Q})$ of order $5$. This element is necessarily a $5$-cycle (easy to see if you think about decomposing the element into disjoint cycles; what's the order of such a decomposition in terms of the lengths of the disjoint cycles?).
Moving on, you've already noted that the polynomial has exactly two complex roots which are necessarily complex conjugates, say $a + bi$ and $a-bi$. There exists a $\phi \in \text{Gal}(K/\mathbb{Q})$, namely complex conjugation, wherein $\phi(a+bi) = a-bi$ and fixes the $3$ real roots. In particular, $\phi$ is a $2$-cycle.
Next, it is a theorem that any $2$-cycle together with any $p$-cycle will generate the entire symmetric group $S_p$ for any prime $p$. From this, we can conclude that $\text{Gal}(K/\mathbb{Q}) \cong S_5$.
All that remains is to show that $S_5$ is not a solvable group. $S_5$ cannot be solvable because $A_5$ is its only normal subgroup, and $A_5$ has no normal subgroups, so we cannot construct a chain $\{e\} = G_0 \subset G_1 \subset \cdots \subset G_n = S_5$ such that each $G_{j-1}$ is normal in $G_j$ and $G_{j}/G_{j-1}$ is abelian.
Therefore, we can conclude the roots of $f$ are not solvable by radicals.
Edit: $ \ $ SteveD in the comments below has brought my attention to Jordan's theorem, which states that, if a subgroup $H \leq S_n$ is primitive and contains a $p$-cycle for a prime $p< n \! - \! 2$, then $H \cong S_n \text{ or } A_n$. One can also show that transitive groups of prime order are primitive, which would allow us to conclude $\text{Gal}(K/\mathbb{Q}) \cong S_n \text{ or } A_n$ after demonstrating the existence of a $5$-cycle or of a $2$-cycle in this group. From here, we are able to conclude that the polynomial cannot be solvable by radicals since neither $S_n$ nor $A_n$ is solvable.
Well, when you derive your polynomial $f$ you get only one real root $1$ for $f'$. Then you see that $f$ is decreasing from $-\infty$ to $1$ and increasing from $1$ to $+\infty$. Now since $f(1)=-1$. We see that $f$ has two real roots and two non-real conjugate complexes.
Now it follows that the coomplex conjugation will fix two roots (the real one) and exchange the two non-real ones. That is the Galois group induces a transposition of the set of roots. It follows that your Galois group as a subgroup of the permutations of the roots $\mathfrak{S}_4$ will contain a transposition. Up to conjugation there are only two transitive subgroups which contain a transposition :
$$\mathfrak{S}_4\text{ and } D_4 $$
Where $D_4$ is one $2$-Sylow of $\mathfrak{S}_4$.
Reducing your polynomial mod $5$ you get $x^4+x+2$. Mod $5$ this factorizes :
$$(x-2)(x^3+2x^2-x-1) $$
Remarking that both polynomials in the factorization are irreducible mod $5$, you have a theorem in your course that say this implies that you have a $3$-cycle in your Galois group. Because of the alternative we already highlighted the Galois group cannot but be $\mathfrak{S}_4$.
Best Answer
\begin{align} \lim_{x\to+\infty}f(x) &= +\infty\\ f(1) &= -1\\ f(0) &= 2\\ \lim_{x\to-\infty}f(x) &= -\infty\\ \end{align} Therefore $f(x)$ has three real roots. You have shown that $f$ has two complex non-real roots, so you found all roots of $f$. Then, by the lemma you have proveded it follows that the Galois group of $f$ is $S_5$ and therefore $f$ is not solvable by radicals over $\mathbb{Q}$.