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EDIT : This about explicit notation and clarity, usage of $\sqrt{(-x)^2}= x$ is for demonstrative purposes not for the problem itself.

$\sqrt {(-2) ^2} = \sqrt {2^2}$ causing confusion $-2=2$ which is nonsense.

What seems to be needed is somehow distinguishing between $-2=2$ and explicitly stating $2,-2$ are in the set $S=$same equivalence class of numbers $k$ s.t. $k^2$ having the same value.

Another observation is the order of operation is important, $\sqrt {(-2)^2} = \sqrt {2^2}$ can be evaluated as either $\sqrt {4} = \sqrt {4} =2$ or $ {-2} = {2}$ , how ever implicitly it known that the square function performs first and only after that the square root function can operate, how could this implicit knowledge be worked into explicit notation?

Are there any notations that can be used to make order of operation explicitly visible or instead of ending up with $2=-2$, somehow end up with $2 \equiv-2$ with respect to their squares having same values?

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    Do you mean $-2^2$ as written or $(-2)^2$? – 2017-02-19
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    @Magdiragdag : Yes fixed, thaks – 2017-02-19
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    If you think that $\sqrt{(-2)^2}=-2$ then its wrong. We know that $\sqrt{a^2}=|a|$ for any real number $a$. – 2017-02-19
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    @ΘΣΦGenSan : no that is not something that we know, that is by definition. – 2017-02-19
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    @Arjang Then why not just use $\pm\sqrt\cdot$? – 2017-02-19
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    @SimplyBeautifulArt : is that a notation improvement? then that is what I am trying to ask for! :) – 2017-02-19
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    Not all functions are injective. we have to live with that. If you fix $\sqrt{x}$ somehow with "better notation", then you will just have the same problem with $\log x$ and with $\arcsin x$. – 2017-02-19
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    So I cast the final vote to close. It was not such an easy decision, for the question pointed to is not really an exact duplicate of this one. However, the (non-accepted) answer to the other question is an excellent answer to the present question, I think. **Edit:** I am referring to the answer by @paulgarrett. – 2017-02-19
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    @GEdgar : Yes, this was a prototype question for $\log , \arcsin $ etc. type that could be be addressed with better notation. – 2017-02-19
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    @HaraldHanche-Olsen : This is about notation , the not accepted answer has nothing to do with giving a better nortation, and it's auther was told about the problem in the comments. – 2017-02-19
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    In my view, the answer has *everything* with the futility of trying to deal with the ambiguity by notational devices. – 2017-02-19
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    @HaraldHanche-Olsen : What other device cane be used to tackle the issues that arise out of ambiguity? I wanted this as a prototype for other similar problems arising with implicit operational precedence and "Inverse" of multivalued functions ending up using "=" sign for different value of parameters that it takes. How else can we be explicit if not by notation? – 2017-02-19
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    You make things clear by explaining what is going on, conveying understanding. And in this particular case, by pointing out that the “rule” $\sqrt{x^2}=x$ does not exist as a general rule, just as a formula that happens to be true when $x\ge0$. Trying to hyperformalize everything tends to add confusion, not clarity. – 2017-02-19

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The problem is that you seem to believe that, for a real number $x$, $\sqrt{x^2} = x$. That is not true in general, as you've already observed. In fact, $\sqrt{x^2} = |x|$.

As for the order of the operations: the notation $\sqrt{x^2}$ makes it abundantly clear that you take the square first and then the square root; without the bar over the $x^2$, i.e., $\sqrt{\phantom{x}\hskip{-4pt}}x^2$, that would be not as clear.

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    No, There is nothing about my belief in the question, asking for clearer notation if any is known – 2017-02-19
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    $\sqrt{x^2} = |x|$ is not a fact, it is by definition. – 2017-02-19
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    oi, as I said it has nothing todo with what I belief, modify your none answer before I believe to modify it. – 2017-02-25
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Note that in $\sqrt{(x)^2}=\sqrt{(-x)^2}$, the square function performs first and only after that the square root function can operate (Implicit operator precedence).

Also, $\sqrt{(x)^2}=\sqrt{(-x)^2}=|x|$, that is, the answer to this is always positive.

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    asking about notation, not answering the question. – 2017-02-19
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    Should have mentioned that in the heading, before asking the question. – 2017-02-19
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    Should have read the question befor answering – 2017-02-19
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    Well, both of us are at fault, mate. Both of us. – 2017-02-19
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    I am stealing your operation precedence part, thanks! – 2017-02-19
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    Haha. Sure, go ahead. – 2017-02-19