Yes.
This is normally expressed in terms of the Newton polygon of the polynomial. Specifically, given an arbitrary field $K$ with a valuation, and a polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, the Newton polygon of $f$ is the lower side of the convex hull of the set of points $(i, v(a_i))$.
If $K$ is algebraically closed, then the valuations of the $n$ roots of $f$ are exactly minus the slopes of the Newton polygon over the $n$ intervals $[0,1], [1,2], \dots, [n-1,n]$.
The most straightforward way I know to check this is to first, completely factor $f$ into linear factors, i.e. $a_n \prod_{i=1}^n (x- \alpha_i)$, with the $\alpha_i$ in order of increasing valuation, write $a_i$ in terms of the $\alpha_j$, and then check that $$v(a_{n-k}) \geq v(a_0) + \sum_{i=1}^k v(\alpha_i)$$ with equality if $v(\alpha_i) < v(\alpha_j)$, so that the Newton polygon is contained in the polygon with slopes $- v(\alpha_1),\dots, -v(\alpha_n)$, and contains the corners of that polygon, hence is equal to that polygon. This inequality and equality can be checked using the definition of valuation.
Similarly, one can check that the tropicalization of a polynomial equation is a piecewise linear function with corners at minus the slopes of the Newton polygon (in fact, it is dual to the Newton polygon in a certain sense, so the edges of the tropicalization correspond to vertices of the Newton polygon and vice versa), and so the solutions of the tropicalization are also minus the slopes of the Newton polygon.
