Integer roots!

Let the roots of the given quadratic equation be $a$ and $b$ .

By Vieta's formula :

$a+b=-7$ and $ab=-14(q^2+1)$

Note that from first condition, if one of the roots is integral , the other one must be integral.

Now, from second condition , we note that $ab<0$ . WLOG let us assume that $a>0$ and $b<0$ . Now set $c=-b$ so that $c>0$ and the new equations become :

$c-a=7$ and $ca=14(q^2+1)$

Let us suppose that both $a$ and $c$ are integers

From first condition they must differ by $7$ . From second equation exactly one of them is a multiple of $7 [(q^2+1)$ can never be a multiple of $7$ ].

But wait, if both of them differ by $7$ and one of them is a multiple of $7$ then other one must also be a multiple of $7$ , which can never happen.

Hence, the given quadratic equation can never have any integral roots.