Roots?

Let $x_ 1$ and $x_ 2$ be the roots of the equation.

We know that $x_ 1$ and $x_ 2$ must satisfy the statements below:

$x_1 + x_2 = -odd$ . This is only possible if: $x_1 = even$ and $x_2 = odd$ .

$x_1 x_2 = odd$ . This is only possible if: $x_1 = odd$ and $x_2 = odd$ .

As there's no integer or rational numbers that satisfy both conditions, we can say that the equation doesn't have rational roots.

Let $x = \frac{p}{q}$ be a rational root to the above quadratic equation. This computes to:

$a(\frac{p}{q})^{2} + b(\frac{p}{q}) + c = 0 \Rightarrow ap^2 + bpq + cq^2 = 0$

If $p,q$ are both odd, then we obtain odd + odd + odd $\neq$ 0. If p is odd(even) and q is even(odd), then we obtain even + even + odd $\neq$ 0. Hence, we only obtain odd integer solutions in both cases so no rational root solutions exist.