#### Answer

$x=3$

#### Work Step by Step

Squaring both sides of the given equation results to
\begin{array}{l}\require{cancel}
\sqrt{2x+3}=x
\\\\
\left( \sqrt{2x+3} \right)^2=(x)^2
\\\\
2x+3=x^2
\\\\
-x^2+2x+3=0
\\\\
-1(-x^2+2x+3)=-1(0)
\\\\
x^2-2x-3=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
x^2-2x-3=0
\\\\
(x-3)(x+1)=0
.\end{array}
Equating each factor to $0$ (Zero Product Property), the solutions are
\begin{array}{l}\require{cancel}
x-3=0
\\\\\text{ OR }\\\\
x+1=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x-3=0
\\\\
x=3
\\\\\text{ OR }\\\\
x+1=0
\\\\
x=-1
.\end{array}
If $x=-1,$ then
\begin{array}{l}\require{cancel}
\sqrt{2x+3}=x
\\\\=
\sqrt{2(-1)+3}=-1
\\\\=
\sqrt{-2+3}=-1
\\\\=
\sqrt{1}=-1
\\\\=
1=-1
\text{ (FALSE)}
.\end{array}
Hence, only $
x=3
$ satisfies the original equation.