When going from $(x+2)^2=5$ to $x+2=\pm \sqrt{5}$, why isn't there also a $\pm(x+2)$?

The question can be rephrased in abstract form as:

If we have an equation of the form $$a^2=b^2$$ why is it equivalent to $a=\pm b$?

Why not $\pm a = \pm b$?

As Dr. Sonnhard Graubner's answer outlined, it can be explained by \begin{align*} &a^2=b^2\\[4pt] \iff\;&a^2-b^2=0\\[4pt] \iff\;&(a-b)(a+b)=0\\[4pt] \iff\;&a-b=0\;\;\;\text{or}\;\;\;a+b=0\\[4pt] \iff\;&a=b\;\;\;\text{or}\;\;\;a=-b\\[4pt] \iff\;&a=\pm b\\[4pt] \end{align*} Thus we have what I'll call the "square-root$\;\pm\;$principle": $$\boxed{ \;\\[4pt] \quad a^2=b^2\;\iff\;a=\pm b\quad \\ } $$ Applying this principle to the problem at hand, we get $$(x+2)^2=5\;\iff\;x+2=\pm\sqrt{5}$$

Hint: Better is to write $$(x+2)^2-\sqrt{5}^2=0$$ and this is, using that $$a^2-b^2=(a+b)(a-b)$$ $$(x+2-\sqrt{5})(x+2+\sqrt{5})=0$$

When solving $(x+2)^2=5$, recall in general that for $x\in\mathbb{R}$ we have $\sqrt{x^2}=|x|$. And since clearly $x+2\in\mathbb{R}$, we have by the last identity that by taking the square root of both sides of $(x+2)^2=5$ $$\sqrt{(x+2)^2}=|x+2|=\sqrt{5}\tag1$$

Then we have reduced the problem to solving $$|x+2|=\sqrt{5}\tag2$$

Recall once more that in general $$|x|=b>0\implies x=b\text{ or }x=-b\tag3$$

Thus putting $(1)$ and $(3)$ together, $$|x+2|=\sqrt{5}\iff x+2=\sqrt{5}\text{ or } x+2=-\sqrt{5}$$

Futhermore, $$x=\sqrt{5}-2\text{ or } x=-\sqrt{5}-2\iff \boxed{x=-2\pm\sqrt{5}}$$