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\begin{document}
\title{How to Solve $x^{2}+x+6\sqrt{x+2}=18$?}
\author{Amit Yaron}
\date{June 15, 2021}

\maketitle
When I see such equations, I like to have a perfect square under the
radical. Here, we can see that $x=2$ is a solution.

How to find others? There are 2 ways: 1 square both sides after one
side contains the radical only, but I don't like squaring a sum of
3 terms. The second is by the substitution:

\[
t=\sqrt{x+2}\Rightarrow t^{2}=x+2\Rightarrow x=t^{2}-2
\]

Now, let us plug it into the equation:

\begin{align*}
(t^{2}-2)^{2}+t^{2}-2+6t & =18\Rightarrow\\
\Rightarrow t^{4}-4t^{2}+4+t^{2}-2+6t=18 & \Rightarrow\\
\Rightarrow t^{4}-3t^{2}+6t+2=18 & \Rightarrow\\
\Rightarrow t^{4}-3t^{2}+6t-16=0 & \Rightarrow\\
\Rightarrow t^{4}-16-(3t^{2}-6t)=0 & \Rightarrow\\
\Rightarrow t^{4}-2^{4}-3t(t-2)=0 & \Rightarrow\\
\Rightarrow(t-2)(t^{3}+2t^{2}+4t+8)-3t(t-2)=0 & \Rightarrow\\
(t-2)(t^{3}+2t^{2}+t+8)=0
\end{align*}

One solution is:

\[
t=2\geq0\Rightarrow x=t^{2}-2=2
\]

Great! But, I cannot find rational roots of the cubic equation:

\[
t^{3}+2t^{2}+t+8=0
\]

But never mind, because 

\[
t=0\Rightarrow t^{3}+2t^{2}+t+8=8
\]

and the polynomial increases where $t\geq0$

So, the polynomial has negative zeros only

and negative numbers are not square roots.

So, the only real solution is:

\[
\boxed{{\color{red}x=2}}
\]

\end{document}