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) $x \in \{1,-2\}$.
 A: It's $$\sqrt{1+2\sqrt{x+3}+x+3}=\frac{x+8}{3}$$ or
$$\sqrt{\left(1+\sqrt{x+3}\right)^2}=\frac{x+8}{3}$$
$$1+\sqrt{x+3}=\frac{x+8}{3}$$ or
$$\sqrt{x+3}=\frac{x+5}{3}$$ and since $x\geq-3$, it's
$$9(x+3)=(x+5)^2,$$
which gives $x=1$ or $x=-2.$.
A: What you did is fine. Now, you can use the rational root theorem in order to find the roots $1$ and $-2$. Since your polynomial is $(x-1)(x+2)(x^2+13x+94)$, there are no more real roots. Note however that you still must check whether or not $-2$ and $1$ are solutions of the original equation.
A: 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. $0;1;2;-1,i,-i$
A: Your question was how to determine the value of $x$ in the given equation: 
$$
\sqrt{x+4+2\sqrt{x+3}}=\frac{x+8}{3}
$$
1.) 
$
x+4+2 \sqrt{x+3} = \bigg( \frac{x+8}{3} \bigg)^2
$
2.) 
$
x+4+2 \sqrt{x+3} =  \frac{x^2+8^2}{3^2}
$
3.) 
$
9(x+4+2 \sqrt{x+3}) =  (x+8)^2
$
4.) 
$
9x+36+18 \sqrt{x+3} =  (x+8)(x+8)
$
5.) 
$
18 \sqrt{x+3}) =  (x^2+16x+64)-9x-36
$
6.) 
$
\sqrt{x+3}) =  \frac{x^2+7x+28}{18}
$
7.) 
$
x+3 =  \bigg( \frac{(x^2+7x+28)({x^2+7x+28})}{18 \times 18} \bigg)
$
8.) 
$
x+3 =  \frac{x^4+14x^3+105x^2+392x+784}{324}
$
9.) 
$
324x+972 =  x^4+14x^3+105x^2+392x+784
$
10.) 
$
x^4+14x^3+105x^2+68x=188
$

The outcome of the equation is that it is a $4^ \text{th}$ degree polynomial of the form
$
a_1x^4+a_2x^3+a_3x^2+a_4x+a_4=0
$
$$
x^4+14x^3+105x^2+68x-188=0
$$
The solution given by my calcuator is:
$
x_1=1,
x_2=-2
$
To find out how that comes about, refer to Quartic polynomial solutions
