No, by a counting argument that's worth remembering: There are $\mathfrak{c}=2^{\aleph_0}$ real numbers, but the set of finite sets of rational numbers is countable. The proof on this last statement is relatively straightforward: since we can map rational numbers to whole numbers, we can map finite sets of rationals to finite sets of whole numbers. But now, to each finite set of whole numbers $\{a_1, a_2, a_3, \ldots, a_n\}$ — which we can obviously assume is given in increasing order — associate the number $2^{a_1}3^{a_2}5^{a_3}\ldots p_n^{a_n}$ where $p_n$ is the $n$th prime. You should be able to convince yourself that this mapping is one-to-one, and so there can only be countably many finite sets of rationals.
In fact, this argument shows that 'almost all' reals have no finite description — whether in terms of rationals, roots of algebraic equations, or even an English description like 'the tenth $x\gt 0$ where $e^x\sin(x) = 1.5$' — at all.