#### Answer

$x=3$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given radical equation, $
x-\sqrt{2x+3}=0
,$ use the properties of equality to isolate the radical. Then square both sides and use concepts of quadratic equations to solve for the variable. Finally, it is a must to do checking of the solution.
$\bf{\text{Solution Details:}}$
Using properties of equality to isolate the radical, the equation above is equivalent to
\begin{array}{l}\require{cancel}
x=\sqrt{2x+3}
.\end{array}
Squaring both sides and using properties of equality to express the result in $ax^2+bx+c=0$ form result to
\begin{array}{l}\require{cancel}
x^2=2x+3
\\\\
x^2-2x-3=0
.\end{array}
In the trinomial expression above, the value of $ac$ is $
1(-3)=-3
$ and the value of $b$ is $
-2
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
1,-3
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
x^2+x-3x-3=0
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x^2+x)-(3x+3)=0
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
x(x+1)-3(x+1)=0
.\end{array}
Factoring the $GCF=
(x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(x+1)(x-3)=0
.\end{array}
Equating each factor to zero (Zero Product Property), then
\begin{array}{l}\require{cancel}
x+1=0
\\\\\text{OR}\\\\
x-3=0
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
x+1=0
\\\\
x=-1
\\\\\text{OR}\\\\
x-3=0
\\\\
x=3
.\end{array}
Upon checking, only $
x=3
$ satisfies the original equation.