ĐK: $x\ge-6$
Ta có:
$4\sqrt{x+6}=x+1$
$\Leftrightarrow \left\{\begin{array}{l}x\ge-1\\16(x+6)=(x+1)^2\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x\ge-1\\x^2-14x-95=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x\ge-1\\\left[\begin{array}{l}x=19\\x=-5\end{array}\right.\end{array}\right.$
$\Leftrightarrow x=19$