#### Answer

$x=3$

#### Work Step by Step

$Given,$ $x=\sqrt (2x+3)$
Squaring both sides,we get:
$x^{2}=2x+3$
$x^{2}-2x-3=0$
$(x-3)(x+1)=0$
$x=3,x=-1$
We need to check by the putting the obtained solutions in
the original equation.
$x=3$
$L.H.S=3$
$R.H.S=\sqrt (2X3+3)=\sqrt 9=3$
$L.H.S=R.H.S$,hence it is a valid solution
$x=-1$
$L.H.S=-1$
$R.H.S=\sqrt (2X-1+3)=1$
$L.H.S$ does not equal $R.H.S$,hence it is an extraneous solution.