Let $\quad f(x)=ax^2+bx+c \quad $ be a quadratic function, $x \in \left (0,d \right )$ a domain of $f(x)$.
If I need $f(x)$ to be positive, that would be achieved for $ x \in \left (-\infty,x_{-} \right) \cup \left (x_{+}, +\infty \right) $.
But what are the ways to check for $x_{-}$ and $x_{+}$ if they belong to the $ \left (0,d \right )$ ?
The question above is an attempt to generalize what should be a solution to a concrete problem: $$\frac{2x^2+x(8-a)-2}{2x(ax+2)}>0, \quad x\in\left (0,-\frac{2}{a} \right ),\quad a<0$$
$$ \frac{2x^2+x(8-a)-2}{2x(ax+2)}= \frac{2 \left(x-\frac{1}{4} \left(a-8-\sqrt{(a-16) a+80}\right)\right) \left(x-\frac{1}{4} \left(a-8+\sqrt{(a-16) a+80}\right)\right)}{2x(ax+2)} $$
Only one of the roots of the numerator belongs to the interval of interest: $$\frac{1}{4} \left(a-8+\sqrt{(a-16) a+80}\right) \in\left (0,-\frac{2}{a} \right ),\quad a<0$$ This holds for any $a<0$.
The inequality $$\frac{2x^2+x(8-a)-2}{2x(ax+2)}>0$$ with $a<0$ holds for $$\left (0,-\frac{2}{a} \right ) \ni x>\frac{1}{4} \left(a-8+\sqrt{(a-16) a+80}\right)$$
To show this mathematically you do as following.
Let $$f(x)=\frac{2x^2+x(8-a)-2}{2x(ax+2)}$$ Find the extremum $$f'(x)= 4x+8-a$$ solve $4x+8-a=0$ for $x$, that is $x=\frac{a-8}4$. Now the question do we have maximum or minimum here. For this we do second derivative test: $$f''(x)=4>0$$ that is we have minimum as you need.
Next, find the roots
$$r_{\pm}(a)=\frac{1}{4} \left(a-8\pm\sqrt{(a-16) a+80}\right) $$ Differentiate with respect to $a$ and solve $r'_{\pm}(a)=0$ you will have no roots, which means $r_{\pm}(a)$ doesn't cross x-axis, so evaluate $r_{\pm}(0)$ to see that $r_{+}(a)>0$.
In order to show it also satisfies $r_{+}(a)<-\frac{2}{a}$ do the same trick with $g(a)=r_{+}(a)+\frac{2}{a}$. That is solve $g'(a)=0$. You should find no real root, therefore all you need is evaluate it at some value of $a$ to discover that $g(a)<0$, which gives exactly what you need,i.e. $r_{+}(a)<-\frac{2}{a}$.