Solving quadratic inequalities I have a quadratic inequality I am halfway through solving but cannot quite figure out the concept.
$$3+\frac{4-x}x>0$$
I so far understand that without knowing whether x is positive, I would need to square it, but I do not know what comes next.

We have
$$3+\frac{4-x}{x} >0$$
$$\frac{3x+4-x}{x} >0$$
$$\frac{2x+4}{x} >0$$
Divide both sides by $2$
$$\frac{x+2}{x} >0$$
Now use wavy curve method
$-\infty+++++ (-2)-----(0)++++++ \infty$ 
As we want L.H.S. to be positive, hence we get
$x \in (-\infty, -2) \cup (0, \infty)$
Edit 1
Let me explain you how I decided the signs.
When $x>0$, then $x+2>0$ and $x>0$, hence $\frac{x+2}{x}$ will be positive.
Again when $x<-2$, then $x+2<0$ and $x<0$, hence $\frac{x+2}{x}$ will be positive.
When $-2<x<0$, then $x+2>0$ but $x<0$, hence $\frac{x+2}{x}$ will be negative.
That is how I decided the sign.
