Solving solution set inequalities problems I need some guidance with solving problems of the following nature:
$$|x^2-3x-1|<3$$
Instinctively, my way of solving it is splitting it into the following equations:
$$x^2-3x-1< 3$$          
$$x^2-3x-1> -3$$
Solving $x^2-3x-1< 3$
$$x^2-3x-4< 0$$ 
$$(x+1)\ (x-4)< 0$$
$$x < -1$$
$$x < 4$$
Solving $x^2-3x-1> -3$
$$x^2-3x-1> -3$$
$$x^2-3x+2> 0$$
$$(x-2)(x-1)> 0$$
$$x > 2$$
$$x > 1$$
So, here $x < -1$ is a valid solution and $x > 1$ is also valid.
In addition, $x < 4$ is valid and $ x > 2$ is also valid.
So we have $-1 > x > 1$ and $2<x<4$.
Firstly, is this possible? Does any $x$ have to satisfy $-1 > x > 1$ and $2<x<4$ simultaneously? Or are these the ranges of allowed inputs? Like, $x$ can be $-2$ despite $2<-2<4$ is not true. 
Secondly, WolframAlpha has the following numberline:

Which means I've not solved this correctly. So how do I solve these properly?

$$|x^2-3x-1|<3\Rightarrow -3<x^2-3x-1<3\\\Rightarrow -3<x^2-2(3/2)x+(3/2)^2-(3/2)^2-1<3\\\Rightarrow 1/4<(x-3/2)^2<25/4$$
Now you have two inequation:


*

*$1/4<(x-3/2)^2\Rightarrow x-3/2<-1/2\space \text{or}\space x-3/2>1/2\Rightarrow x<1\space \text{or}\space x>2$.

*$(x-3/2)^2<25/4\Rightarrow -5/2<x-3/2<5/2\Rightarrow x>-1\space \text{or} \space x<4$
Hence the final answer is $\{x\in\mathbb{R}: -1<x<1\space\text{and}\space 2<x<4\}\space\space\space\blacksquare$
