When we wish to solve a differential equation like your example, a good guess for when $f(t)=0$ is something of the form $x(t) = e^{i \alpha t}$. This just gives us an algebraic equation for $\alpha$, which we can just solve. I interpret your question to mean: but what is this solution? Why is it complex? Indeed, position should certainly be just a real number. The resolution is that a general solution to the differential equation will not be of the form $x(t)=e^{i\alpha t}$ for $\alpha$ solving our algebraic equation. In general there will be a couple solutions to the algebraic equation for $\alpha$--just two in your example--say $\alpha_1$ and $\alpha_2$. Then a general solution to your diffeq with $f(t)=0$ is of the form $x(t) = A e^{i \alpha_1 t} + B e^{i \alpha_2 t}$ for some complex constants $A$ and $B$. We then need boundary conditions to choose $A$ and $B$ properly. The crux of the biscuit is we can choose $A$ and $B$ so that an $x(t)$ of this form is real for all times $t$ (this assumes $a$, $b$, and $c$ are real in the original equation).
So using the complex guess $x(t) = e^{i\alpha t}$ is just a computational trick which makes things easy because the complex numbers have nice properties with algebraic equations (they are algebraically closed). We could have done the whole thing without $e$, instead just using sines and cosines and it would have worked out too.