Solve the following equation:

$$
\sqrt{4 x^{2}+4 x+1}-\sqrt{4 x^{2}-12 x+9}=4 .
$$
We can immediately see that there are perfect squares under the square root. We know that $\sqrt{a^{2}}=|a|$, thus we arrive at the following equation:

$|2 x+1|-|2 x-3|=4$

According to the definition of the absolute value function,

$$
\begin{aligned}
& |2 x+1|=\left\{\begin{array}{r}
2 x+1, \text { if } x \geqq-\frac{1}{2} \\
-(2 x+1), \text { if } x<-\frac{1}{2}
\end{array}\right. \\
& |2 x-3|=\left\{\begin{array}{r}
2 x-3, \text { if } x \geqq \frac{3}{2} \\
-(2 x-3), \text { if } x<\frac{3}{2}
\end{array}\right.
\end{aligned}
$$

Therefore, we need to examine the solution of the equation in 3 intervals.

1. If $x \geqq \frac{3}{2}$, then $2 x+1-(2 x-3)=4$,

which results in the identity equation $4=4$. Thus, these values of $x$ are solutions to the equation.

2. If $-\frac{1}{2} \leqq x<\frac{3}{2}$, then $2 x+1-[-(2 x-3)]=4$.

From this, $x=\frac{3}{2}$, but this value is not in the interval.

3. If $x<-\frac{1}{2}$, then $-(2 x+1)-[-(2 x-3)]=4$, which results in the contradiction equation $-4=4$.

Thus, there are no solutions to the equation in this interval.

Of the three resulting equations, the first is an identity, and the third is a contradiction.

The solutions to the equation are the numbers on the half-line $x \geqq \frac{3}{2}$.