(Since you're discussing minimal polynomials and algebraic numbers, I'm going to assume that you have some familiarity with Galois theory. If not, this solution might not be too helpful). The computation that I think you're interested in is as follows: In general, if you have some algebraic number, $\alpha$, you can find some Galois extension $K \supset \mathbb{Q}$ with $\alpha \in K$. Then the minimal polynomial of $\alpha$ is $m_\alpha(x) = \prod_{i} (x - \alpha_i)$ where the $\alpha_i$ are the distinct Galois conjugates of $\alpha$. The reasons this computation works are as follows:
You can see that the polynomial is guaranteed to have coefficients in $\mathbb{Q}$ because applying any element of $\operatorname{Gal}(K / \mathbb{Q})$ to $m_\alpha(x)$ will give the same polynomial. Hence, the coeffients of $m_\alpha(x)$ are fixed by all of the automorphisms in $\operatorname{Gal}(K / \mathbb{Q})$, implying that they are all in $\mathbb{Q}$.
Further, you can see the minimality of the degree of $m_\alpha$ with the following argument. Suppose that $f(x) \in \mathbb{Q}[x]$ has $f(\alpha) = 0$. Then for any $\sigma \in \operatorname{Gal}(K/\mathbb{Q})$, we have $f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0$, implying that every distinct Galois conjugate of $\alpha$ is a root of $f$. Hence, $m_\alpha(x)$ divides $f(x)$, which gives the minimality of its degree.
For the example you give above with $\alpha = \frac{3 + \sqrt{-5}}{7}$, we see that $\alpha \in \mathbb{Q}(\sqrt{-5})$, which has two members of its Galois group: the identity and the automorphism which fixes $\mathbb{Q}$ and sends $\sqrt{-5}$ to $-\sqrt{-5}$. Hence, the Galois conjugates of $\alpha$ are $\alpha$ itself and $\overline{\alpha} = \frac{3-\sqrt{-5}}{7}$. Then the minimal polynomial of $\alpha$ is given by $$m_\alpha(x) = (x - \alpha)(x - \overline{\alpha}) = x^2 - \frac{6}{7}x + \frac{2}{7}$$