#### Answer

$x=-1$

#### Work Step by Step

Square both sides:
$(\sqrt{x+2}+1)^2=(\sqrt{2x+6})^2
\\(\sqrt{x+2})^2+2(\sqrt{x+2})(1)+1^2=2x+6
\\x+2+2\sqrt{x+2}+1=2x+6
\\2\sqrt{x+2}+x+3=2x+6$
Subtract $x$ and $3$ to both sides of the equation to obtain:
\begin{array}{ccc}
&2\sqrt{x+2}+x+3-x-3 &= &2x+6-x-3
\\&2\sqrt{x+2} &= &x+3
\end{array}
Square both sides of the equation to obtain:
$$\left(2\sqrt{x+2}\right)^2=(x+3)^2
\\4(x+2)=x^2+2(x)(3)+3^2
\\4(x)+4(2)=x^2+6x+9
\\4x+8=x^2+6x+9$$
Subtract $4x$ and $8$ to both sides of the equation to obtain:
\begin{array}{ccc}
&4x+8-4x-8 &= &x^2+6x+9-4x-8
\\&0 &= &x^2+2x+1
\end{array}
Factor the trinomial to obtain:
$$0=(x+1)(x+1)$$
Use the Zero-Factor Property by equating each factor to zero, then solve each equation to obtain:
$x+1=0$ or $x+1=0$
$x=-1$ or $x=-1$
Thus, the solutions is $x=-1$.