Different solutions with different results for an inequality Find m such that the following inequality:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
is always true for $\forall x \in R$.
1st solution:
1st case
$$4x-2m-\frac{1}{2} > -x^2 + 2x +\frac{1}{2} -m$$
$$<=>x^2+2x-m-1>0$$
$$\Leftrightarrow 1^2+(m+1)< 0$$
$$\Leftrightarrow m<- 2$$
2nd case
$$4x-2m-\frac{1}{2}< -(-x^2 + 2x +\frac{1}{2} -m)$$
$$\Leftrightarrow x^2-6x+3m>0$$
$$\Leftrightarrow 3^2-3m<0$$
$$\Leftrightarrow m>3$$
2nd solution:
The inequality is the same as:
$$(x-1)^2+|4x-2m-\frac{1}{2}|>\frac{3}{2}-m$$
Since the left-hand side is always positive, in order for the inequality to be always true, $\frac{3}{2}-m$ has to be negative, or $m > \frac{3}{2}$

The 2 solutions give different answers, so I was quite confused
But I get more confused as Wolfram Alpha gives me the solution:
$$m > \sqrt{3} - \frac{1}{4} \text{  or  } m < -\sqrt{3} - \frac{1}{4} $$
There's a high chance that Wolfram Alpha's solution is correct (after testing out some $m$ value). How do I approach their solution? (Or maybe if you believe that solution is wrong, then what's the exact solution to the problem?)

Your two solutions are mathematically correct, but you when you reach $m<-2$, you have calculated for which values of $m$, we have $m+2>0$ and you have not shown that $\forall m>-2, (x+1)^2>m+2, \forall x \in R$. This explain the error in the first method. For the second method, you have shown that the inequality:
$$(x-1)^2+|4x-2m-\frac{1}{2}|>\frac{3}{2}-m$$
have solutions when $m<\frac{3}{2}$, but this not explain for which $m$, we have:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
In general, when you have to solve this type of inequality, you have to solve the system:
$$\left\{\begin{matrix}
4x-2m-\frac{1}{2}\geq0
\\ -x^2 +2x + \frac{1}{2} - m\geq0
\\ 4x-2m-\frac{1}{2} > -x^2 +2x + \frac{1}{2} - m
\end{matrix}\right.
\vee
\left\{\begin{matrix}
4x-2m-\frac{1}{2}<0
\\ -x^2 +2x + \frac{1}{2} - m<0
\\ -(4x-2m-\frac{1}{2}) > -x^2 +2x + \frac{1}{2} - m
\end{matrix}\right.
$$
These system are quite complicated, and I will give directly the solutions:
$$\left\{\begin{matrix}
x>\sqrt{m+2}-1
\\ x<-\sqrt3\sqrt{3-m}+3
\end{matrix}\right.$$
When $m<-2$ and $m>3$ the square roots are not defined, so these values of $m$ are correct. 
Also, we want the two expression to be the same because we want a single solution, so:
$$\sqrt{m+2}-1=-\sqrt3\sqrt{3-m}+3$$
The solutions are:
$$m=\sqrt3-\frac{1}{4} \vee m=-\sqrt3-\frac{1}{4}$$
When $-\sqrt3-\frac{1}{4}<m<\sqrt3-\frac{1}{4}$ there are solutions, while when:
$$m > \sqrt{3} - \frac{1}{4} \lor m < -\sqrt{3} - \frac{1}{4}$$
there aren't.
