Clearly, $c\neq 0$, as $f$ has no real roots. Now, note that, ${\rm deg}(g-h)=1$ (since, their leading coefficients are the same, and the coefficients of $x$ are different). Thus, there is an $x_0$, such that $g(x_0)=h(x_0)\neq 0$ holds. Then, it suffices to let $c_0=-f(x_0)/g(x_0)$. The polynomials $f+c_0 g$ and $f+c_0 h$ both have $x_0$ as their common root.