Find all values of $c$ such that solutions of $x^2 + cx + 6 = 0$ are rational. This is what I've done so far:
Using the formula, we get
$x = \frac{-c \pm \sqrt{c^2 - 24}}{2}$
For solutions to exist, $c \geq 5$.
For solutions to be rational, $c^2 - 24$ has to be a perfect square.
I'm kind of stuck now. I'd appreciate if someone can give me a hint or two on how to solve this question. Thanks!

Observe that if $x \in \mathbb{Q}$ which is a solution of the equation $x^2 + cx + 6 = 0\text{ } (1)$, then solve for $c$ in terms of $x$ we obtain: $c = -\dfrac{x^2+6}{x}$. Thus if $c = - \dfrac{r^2+6}{r}$ with $r \in \mathbb{Q}\setminus \{0\}$, we prove the solutions of $(1)$ are rational numbers. We have: $x^2+ cx + 6 = 0 \implies x^2 - \left(\dfrac{r^2+6}{r}\right)x + 6 = 0\implies x^2r-r^2x - 6(x-r) = 0\implies (xr-6)(x-r)=0\implies x = r, \dfrac{6}{r}$ which are both rationals.
