#### Answer

The solution of the given equation is $\frac{1}{2}$.

#### Work Step by Step

Square both sides to obtain:
$(4x)^2=(\sqrt{2x+3})^2
\\16x^2=2x+3
\\16x^2-2x-3=0$
Factor the trinomial:
$(8x+3)(2x-1)=0$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
\\&8x+3=0 &\text{or} &2x-1=0
\\&8x=-3 &\text{or} &2x=1
\\&x=-\frac{3}{8} &\text{or} &x=\frac{1}{2}
\end{array}
Note that when $x=-\frac{3}{8}$: the left side of the given equation becomes $-\frac{3}{2}$.
Since the right side of the given equation involves a principal square root, its value must be non-negative.
Thus, $-\frac{3}{8}$ is not a solution of the equation.
Therefore, the solution of the given equation is $\frac{1}{2}$.