The given equation is
$$x^2-x+a-3<0$$
And
$$x^2 + x -(a+3) <0$$

The right answer is 6, which can be obtained by replacing the inequity with an equality, making the discriminant positive and adjusting for a negative root to obtain the interval of $a$

$$x^2-x+a-3=0$$
And
$$x^2+x-(a+3)=0$$

So $a>-\frac{13}{4}$ and $a< \frac{13}{4}$

Also $$x=\frac{1\pm \sqrt{13-4a}}{2}$$
So $$\sqrt{13-4a}>1$$
$$a<3$$

That gives $a\in (-\frac{13}{4}, 3)$ ie. 6 integral values

But this method doesn’t sit right with me, since values satisfying the inequality were supposed to be found, not the equality. Maybe it’s just a technicality, but I am not convinced. Can I get an explanation for this?

*I know a similar question already exists in the site, but I found it unsatisfactory and it doesn’t answer my specific problem*