<s>[INST] When going from $(x+2)^2=5$ to $x+2=\pm \sqrt{5}$, why isn't there also a $\pm(x+2)$? Say I am solving the following equation:
$$(x+2)^2 = 5$$
$$x + 2 = \pm \sqrt{5}$$
$$x = -2 \pm \sqrt{5}$$
However, when I took the positive and negative square root of $5$ in the second line, I did not take the positive and negative square root of $(x+2)^2$, which would be $\pm (x+2)$. Why is this?
 [/INST] Hint: Better is to write
$$(x+2)^2-\sqrt{5}^2=0$$ and this is, using that $$a^2-b^2=(a+b)(a-b)$$
$$(x+2-\sqrt{5})(x+2+\sqrt{5})=0$$
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