This thread is a couple of months old, but I think I should weigh in, given that the other answers seem to focus on the solutions of algebraic equations.

I think it is worth stressing that, despite what the examples given might make you think, the concept of "gaps" is inherently topological (or order-theoretic, whichever way you prefer to swing), not algebraic.

Indeed, what happens is sort of the opposite of what the example with the square root seems to suggest: what we do is we fill the gaps (most of which we cannot even name!). *Then*, having done that, we see that, lo and behold, we have e.g. a positive $n$-th root for each positive number --- but this can be seen as a sort of side effect. Merely ensuring that we have all these solutions will *not* give us completeness --- the magic only works in one way.

In other words, even if you extend the rationals to a larger ordered field to ensure that the resulting field contains solutions for all algebraic (or even analytic!) equations (which admit real/ordered solutions at all, i.e. not including equations like $x^2+1=0$, or "too many" solutions of equations like $x^2+x+1=0$), the resulting field will usually not be complete (i.e. "gap-free"), and depending on how you choose to extend the order$^\dagger$, it might not even be a subfield of the real numbers (e.g. it might contain infinitesimals).

The easy way to see this is by noting that there are only countably many equations (in finitely many variables, at least), so you can start with the rational numbers, take all these equations with rational parameters plugged in, add solutions, generate a field $F_1$ with those, then take all equations with parameters in $F_1$, use their solutions to obtain a field $F_2$, rinse and repeat, and then $\bigcup_n F_n$ will be a countable field in which all possible equations have solutions. It not very difficult to see that a countable dense linear order cannot be complete (using the fact that the completion of rationals has the cardinality of the continuum). All this remains true even if you add equations using e.g. exponentials, trigonometric functions, integrals, etc.

(That is of course, unless you allow external parameters --- if you add all solutions to equations of the form $x=r$ where $r$ is a real number, then the resulting set will certainly contain the reals, and if you add nothing else besides, you will, of course, get the set of real numbers.)

Even if you add some continuum many elements extra on top of that (and *then* ensure all equations have solutions), there is no reason for what you get to turn out complete.

$\dagger$ Actually, once you have the solutions to all equations of the form $x^2-y=0$ with $y\geq 0$, it is easy to see that there is a unique total ordering compatible with multiplication, namely that for which the positive elements are exactly the squares. That is, provided you are careful enough not to add nonzero $x,y$ with $x^2=-y^2$. The way I am stating this here, this is a bit circular, but it can all be stated and shown in a formally sound way which I do not want to get into to avoid being too technical.