Find the range of $a$ for which the quadratic expression $ax^2+(a-2)x-2$ is negative for exactly two INTEGRAL values of $x$.
The official answer is $a\in [1,2)$
My approach is as follow
The value $(a-2)^2+8a\ge 0$, hence no complex root.
$a>0$ is other condition
$f(n)=an^2+(a-2)n-2>0$
$f(n+1)=a(n+1)^2+(a-2)(n+1)-2<0$
$f(n+2)=a(n+2)^2+(a-2)(n+2)-2<0$
$f(n+3)=a(n+3)^2+(a-2)(n+3)-2>0$
Not able to approach from here
I don't see any particularly easy way to use your approach. Instead, note we can factor the quadratic expression to get that we're looking for where there's exactly $2$ integral values of $x$ satisfying
$$f(x) = ax^2 + ax - 2x - 2 = ax(x+1)-2(x+1) = (ax-2)(x+1) \lt 0 \tag{1}\label{eq1A}$$
As you've already stated, $a \gt 0$ so the parabola is concave up (otherwise, for $a = 0$, it's a straight, non-horizontal line, and for $a \lt 0$ it's concave down, with there being infinitely many integral values of $x$ where $f(x)$ is negative in both cases). One of the factors in \eqref{eq1A} must be negative while the other one is positive. If $x + 1 \lt 0 \;\to\; x \lt -1$ then, since $a \gt 0$, we have $ax - 2 \lt 0$ as well, so this is not allowed. Thus, we must have
$$x + 1 \gt 0 \;\to\; x \gt -1, \;\;\; ax - 2 \lt 0 \tag{2}\label{eq2A}$$
The first condition means that the integral values of $x$ we're looking for must be $0$ and $1$, so the second condition must hold for those $2$ values, but with $x = 2$ not working, so the second condition must not be true then. For $x = 0$, we have $-2 \lt 0$, while for $x = 1$ we require $a - 2 \lt 0 \;\to\; a \lt 2$. Finally, for $x = 2$, we need $2a - 2 \ge 0 \;\to\; a \ge 1$. Putting these together gives the official answer you provided, i.e.,
$$1 \le a \lt 2 \;\to\; a \in [1,2) \tag{3}\label{eq3A}$$
Alternatively, from \eqref{eq1A}, $f(0) = -2$ means $0$ is one of the integral values of interest. Since $f(x)$ must have $2$ roots, and $f(-1) = 0$, the other root is $1 \lt x \le 2$. Thus, then $f(1) = 2a - 4 \lt 0$, and $f(2) = 6a - 6 \ge 0$. This becomes
$$2a - 4 \lt 0 \;\to\; a \lt 2, \;\;\; 6a - 6 \ge 0 \;\to\; a \ge 1 \tag{4}\label{eq4A}$$
These $2$ conditions together then also leads to \eqref{eq3A}.