Example 7 Solve the equation $\sqrt{4 x+3}+2 \sqrt{1-2 x}=\sqrt{15}$.
Solve $15=\left(\sqrt{2} \cdot \sqrt{2 x+\frac{3}{2}}+2 \cdot \sqrt{1-2 x}\right)^{2}$
$$\begin{array}{l}
\leqslant\left[(\sqrt{2})^{2}+2^{2}\right] \cdot\left[\left(\sqrt{2 x+\frac{3}{2}}\right)^{2}+(\sqrt{1-2 x})^{2}\right] \\
\leqslant 6\left(2 x+\frac{3}{2}+1-2 x\right)=6 \times \frac{5}{2}=15
\end{array}$$

The necessary and sufficient condition for the equality to hold is: $\frac{\sqrt{2 x+\frac{3}{2}}}{\sqrt{2}}=\frac{\sqrt{1-2 x}}{2}$, solving this yields $x=-\frac{1}{3}$.

Therefore, the root of the original equation is $x=-\frac{1}{3}$.