<s>[INST] What is $x$ if $\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$? I need to find $x$, given that
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
I simplified this to $x^4+14x^3+105x^2+68x-188=0$. According to Symbolab, that is not correct. That's why I'm goint to write out my attempt so you can point out where my mistake is.

My attempt
$$\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}$$
$$\left(\sqrt{x+4+2\sqrt{x+3}}\right)^2=\left(\frac{x+8}{3}\right)^2$$
$$x+4+2\sqrt{x+3}=\frac{x^2+16x+64}{9}$$
$$9x+36+18\sqrt{x+3}=x^2+16x+64$$
$$-x^2-7x-28=-18\sqrt{x+3}$$
$$x^2+7x+28=18\sqrt{x+3}$$
$$(x^2+7x+28)^2=(18\sqrt{x+3})^2$$
$$x^4+7x^3+28x^2+7x^3+49x^2+196x+28x^2+196x+784=324x+972$$
$$x^4+14x^3+105x^2+68x-188=0$$
Where is my mistake? Even if this were true, I still wouldn't be able to solve it without a calculator (I can't use Rational Root Theorem on such a big numbers!).

By the way, the solution should be (again, according to Symbolab) $x \in \{1,-2\}$.
 [/INST] You can brutally use Ferrari formula but no need here.
Search for obvious Roots , here 1 is clearly an obvious Roots then factorise by :
$$X-1$$
And try again with obvious roots. $0;1;2;-1,i,-i$
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