Do $c=\frac{3\pm\sqrt{13}}2$ have meaningful 2-adic valuations? Do $c=\dfrac{3\pm\sqrt{13}}2$ have meaningful 2-adic valuations?
If not, is there an extension to $\lvert\cdot\rvert_2$ such that either
a) they do, or
b) such that $\lvert c\rvert_p=1/c$?
I don't know where to start with irrational padic values.
 A: Yes, they do. In spite of the problem in GEdgar's comment.
The extension $K=\Bbb{Q}_2(\sqrt{-3})$, also gotten by adjoining a primitive third root of unity $\omega=(-1+\sqrt{-3})/2,$ is a so called unramified extension. Implying (among other things) that:


*

*The ring $R=\Bbb{Z}_2(\omega)$ is the integral closure $\Bbb{Z}_2$ in $K$.

*$R$ is a local ring.

*The principal ideal $2R$ generated by $2$ is the unique maximal ideal of $R$ consisting of its non-units.

*The quotient ring $R/2R$ is isomorphic to the field of four elements $\Bbb{F}_4$, cosets represented by $0,1,\omega$ and $\omega+1\equiv\omega^2$.


The connection to your question is explained by the following:


*

*$-39\equiv1\pmod8$ so the field $\Bbb{Q}_2$ contains square roots of $-39$.

*Because $13=(-39)/(-3)$, the field $K$ contains $\pm\sqrt{13}$.

*The $2$-adic valuation extends to $K$: Every non-zero element $z\in K$ can be written in the form $z=2^mu$ with $u\in R^*=R\setminus2R$, and the exponent $m\in\Bbb{Z}$ then describes the $2$-adic valuation as usual. See also here for a description of that extension of the valuation using the norm map $N:K\to\Bbb{Q}_2$.

*The irreducible quadratic $x^2-3x-1\in\Bbb{Z}_2[x]$ has a constant term in $\Bbb{Z}_2^*$ implying that its roots are units of the ring $R$. Therefore their extended values are $|(3\pm\sqrt{13})/2|=1$. Or, if you prefer the exponential form, $\nu(c)=0$.

*Whether $\dfrac{3+\sqrt{13}}2$ is congruent to $\omega$ or $\omega+1$ modulo $2R$ depends on the choices you make when specifying the square roots (this is built into GEdgar's comment).


Worth knowing:


*

*Possibly surprisingly $\Bbb{Q}_2$ has only finitely many quadratic extensions. All listed locally here. In Pete L. Clark's list $\Bbb{Q}_2(\sqrt5)=\Bbb{Q}(\sqrt{-3})$ is this (unique) unramified extension, due to the fact that $3\cdot(-5)=-15\equiv1\pmod 8$, so $\sqrt{-15}\in\Bbb{Q}_2$.

*The other quadratic extensions are ramified, meaning that the element $2$ no longer generates a maximal ideal in the integral closure $\Bbb{Z}_2$ with the other extension fields.

A: This is to amend Jyrki's answer with one warning and one generalisation from a different perspective.
Warning: Everyone remembers from high school that the two solutions of $x^2-3x-1$ are $x_{1,2}=\frac12 (3 \pm \sqrt{13})$, where the number $13 := 3^2-4\cdot(-1)$ is the discriminant of the polynomial. However, in high school it's tacitly assumed that that symbol $\sqrt{13}$ means the square root $\color{red}{\textit{in} \; \Bbb R}$. If we look at some other extension $K$ of $\Bbb Q$, the formula for the solutions still is correct, but now we have to interpret that square root of $13$ as a square root $\color{red}{\textit{in} \; K}$ (if such a square root exists). I stress this because the OP has a history of mixing up real numbers with $p$-adic numbers: It's not the real numbers $x_{1, real} \approx 3.3027756377...$ and $x_{2, real} \approx -0.3027756377...$ which get a $2$-adic value in Jyrki's answer. It is two elements of $\Bbb Q_2(\omega)$, each of which one could write down as 
$$x_{1, \Bbb Q_2(\omega)} = (\text{some binary string}) + (\text{some other binary string})\cdot \omega$$
$$x_{2, \Bbb Q_2(\omega)} = (\text{some other binary string}) + (\text{some other binary string})\cdot \omega,$$
and it is these two numbers which have well-defined $2$-adic values. As long as one does only algebra, it is OK to just write $x_1 = \frac12(3+\sqrt{13})$ both for the real number and the $2$-adic number (to quickly compute their powers etc.). But as soon as one is interested in ordering, topology, "which of $x_1, x_2$ is closer to $5$?", ..., one should actually not express either the real or the $2$-adic numbers with that algebraic formula, but needs at least a little bit of their actual real, resp. $2$-adic, expansions.
(By the way, another way to express those $2$-adic solutions would be as Witt vectors over the finite field $\Bbb F_{2^2}$.)
Generalisation: Let $p$ be any prime and let $x^2+bx+c$ be a monic quadratic polynomial $\in \Bbb Q[x]$. Everyone remembers from high school that the two solutions are $x_{1,2}=\frac12 (-b \pm \sqrt{D})$, where the number $D := b^2-4c$ is the discriminant of the polynomial, and in high school $\sqrt{D}$ means the square root in $\Bbb R$, if it exists (or in cool high schools, the square root in $\Bbb C$, where it is guaranteed to exist).
Of course if $D$ is a square in $\Bbb Q$, the polynomial splits over $\Bbb Q$; if not, it might happen that $D$ is a square in $\Bbb Q_p$, or if not, then there is a quadratic extension $K$ of $\Bbb Q_p$ which contains a square root of $D$ and hence the polynomial splits (your example falls in this third situation).
Either way, the solutions $x_1$, $x_2$ have well-defined $p$-adic values, and it turns out one can read them off the values of the coefficients $b$ and $c$. (This can be done with Vieta's formulae, or algebraic number theory, or most easily, with Newton polygons.)
It turns out that from this perspective there are just two cases:

Case I: If $v_p(b) \ge \frac12 v_p(c)$, then both solutions of $x^2+bx+c=0$ have the same value
$$\lvert x_i \rvert_p = \lvert c \lvert_p^{1/2}.$$
Case II: If $v_p(b) < \frac12 v_p(c)$, then one of the solutions has the same value as $b$, and the other the same as $c/b$:  $$\lbrace \lvert x_1 \rvert_p, \lvert x_2 \rvert_p \rbrace = \lbrace \lvert b \rvert_p, \lvert \frac{c}{b} \rvert_p \rbrace$$

Further, one can show that if the polynomial is irreducible over $\Bbb Q_p$, we are in case $I$, i.e. we have $$\lvert x_i \rvert_p = \lvert c \lvert_p^{1/2}.$$
The interesting tricky case is that the polynomial is irreducible over $\Bbb Q$, but splits over $\Bbb Q_p$, and one is in case II -- user mercio brought up an example for this when I was too sloppy here: For the polynomial $x^2-x+2$, one is tempted to write the two solutions as $x_{1,2} = \frac12( 1\pm\sqrt{-7})$, but that does not help much, because, just like in user GEdgar's comment, there are two numbers in $\Bbb Q_2$ whose square is $-7$; and now it turns out that we are in case II, one of the $x_i$ has $2$-adic value $\lvert x_i \rvert_2 = 1$, and the other has value $\lvert x_i \rvert_2 = 1/2$ -- but which is which in the $\pm$ formula? That question makes no sense, because it's unclear which of the two square roots of $7$ in $\Bbb Q_2$ the "$\sqrt{-7}$" in the formula refers to. What Jyrki and I figured out there was:

If by $\sqrt{-7}$ we mean the $2$-adic number which ends in $...11$, then $\lvert \frac12( 1 + \sqrt{-7}) \rvert_2 = 1/2$ and $\lvert \frac12( 1 - \sqrt{-7}) \rvert_2 = 1$;
but if by $\sqrt{-7}$ we mean the $2$-adic number which ends in $...01$, then $\lvert \frac12( 1 + \sqrt{-7}) \rvert_2 = 1$ and $\lvert \frac12( 1 - \sqrt{-7}) \rvert_2 = 1/2$.

