Đặt căn x=a
=>\(\sqrt{3+2a}+a=6\)
\(\Leftrightarrow\sqrt{2a+3}=6-a\)
\(\Leftrightarrow\left\{{}\begin{matrix}a< =6\\a^2-12a+36=2a+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a< =6\\a^2-14a+33=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a< =6\\\left(a-3\right)\left(a-11\right)=0\end{matrix}\right.\Leftrightarrow a=3\)
=>x=9