I assume you are asking how to solve inequalities by squaring both sides in general regardless of this particular example.
You can consider this as a general rule:
You can never square both sides of an inequality, unless both sides are positive.
You already noticed why this rule must be followed, when you tried to square both sides of this inequality:
$$2 > -3$$
So when you're asking: "When to flip the inequality sign?". The answer is:
You never flip the inequality sign, when squaring both sides in accordance with the aforementioned rule.
Then you may ask: "How do I solve an inequality which contains a square root (needs squaring in order to solve it), and which does not have 2 positive sides?"
There are 3 elementary cases for an inequality with a square root:
- Both sides are positive:
Square both sides of the inequality, and continue accordingly, ex:
\begin{align*}
5 &> \sqrt{-x} &&\text{:Defined only when } x\leq0\\
25 &> -x &&\text{:Square both sides}\\
x &> -25 &&\text{:The Pre-Solution}\\
-25 &< x \leq 0 &&\text{:Final solution is the intersection of the definition and the pre-solution}\\
\end{align*}
- Both sides are negative:
Multiply both sides by $-1$ (and flip the inequality sign), and it's reduced to case 1. Ex:
\begin{align*}
-5 &< -4\sqrt{1+x^2} &&\text{:Defined when } x \in \mathbb{R}\\
5 &> 4\sqrt{1+x^2} &&\text{:Multiply both sides by }-1\\
25 &> 16(1+x^2) &&\text{:Square both sides}\\
9 &> 16x^2\\
x^2 &< \frac{9}{16}\\
-\frac{3}{4} < x &< \frac{3}{4} &&\text{:Pre-solution}\\
x \in &\left(-\frac{3}{4},\frac{3}{4}\right) &&\text{:Final solution}
\end{align*}
- One side is positive and one is negative:
Here the inequality is either impossible or always true, because positive is always greater than negative, ex:
\begin{align*}
-3&<\sqrt{x}&&\text{:Defined when }x\geq0 \text{ and always true when }x\geq0\\
5&<-\sqrt{1-x^2}&&\text{:Impossible}
\end{align*}
P.S, When you have a side that is not entirely positive or negative, you divide the domain of definition into parts where it is either entirely positive or entirely negative, to reduce the problem to one or more of the cases above. Ex:
$$
x > \sqrt{x+6} \quad \text{:Defined when }x \geq -6\\
$$
Here, LHS is negative when $x \in [-6,0)$, and positive when $x \in [0,+\infty)$.
So, when $x \in [-6,0)$ we have case 3, and the inequality is impossible.
When $x \in [0,+\infty)$ both sides are positive and we have case 1, so we can square both sides:
\begin{align*}
x^2 &> x+6 && \text{:Square both sides}\\
x^2-x-6 &> 0\\
\end{align*}
\begin{align*}
&x \in (-\infty,-2) \cup (3,+\infty) &&\text{:Pre-solution}\\
&x \in (3,+\infty) &&\text{:Final solution is the intersection of the Pre-solution with }[0,+\infty)
\end{align*}
Now, to address your example:
$$x>-2\sqrt{2x-4} \qquad \text{:Defined when }x \geq 2$$
Here, when $x \geq 2$ LHS is positive, and RHS is negative. So this is case 3 (always true when defined).
So this inequality is always true when $x \geq 2$. (No need to square both sides)
$$x \in [2,+\infty) \quad \text{:Final Solution}$$
In 1., $x^2-4\ge0$ is not equivalent to "x has to be greater equal +2 and -2" but to $x\in L_1,$ where
$$L_1=(-\infty,-2]\cup[2,+\infty).$$
In 2., the notations $x<\pm2$ and $x>\pm1$ are similarly confusing and incorrect, but the conclusion is ok:
$$L_2 = (-2,-1) \cup (1,2).$$
Your final solution would anyway be false, since $[2,\infty)\cup(-2,-1)\cup(1,2)$ is not equal to $(-2,-1)\cup(2,\infty)$ but to $(-2,-1)\cup(1,\infty).$
Now, the correct solution is indeed
$$L=L_1\cup L_2=(-\infty,-1) \cup (1,\infty).$$
Note that we knew in advance that each of the sets $L_1,L_2,L$ would be symmetric with respect to $0,$ since they are given by inequations on $x^2.$
Best Answer
Caution,
$$a^2>b$$ does not imply $$a>\pm\sqrt b.$$
But $$\pm a>\sqrt b$$ is correct. (With a somewhat sloppy notation.)
More rigorously
$$a^2>b\\\iff (a-\sqrt b)(a+\sqrt b)>0\\\iff (a>\sqrt b\land a>-\sqrt b)\lor (a<\sqrt b\land a<-\sqrt b)\\\iff a>\sqrt b\lor a<-\sqrt b.$$