Solve the following equation:

$$
(x-2)^{2}+\left(|x-1|+|x-3|-\frac{15}{4}\right)^{2}=\frac{65}{16}
$$
According to the definition of absolute value, we examine three cases.

i) If $x \leq 1$. Then $|x-1|=-(x-1)$ and $|x-3|=-(x-3)$.

The starting equation after removing the absolute value signs is:

$$
x^{2}-4 x+4+\left(-x+1-x+3-\frac{15}{4}\right)^{2}=\frac{65}{16}
$$

After rearranging, we get $5 x^{2}-5 x=5 x(x-1)=0$. From this, either $x=0$ or $x=1$.

ii) If $1 < x \leq 3$, then $|x-1|=x-1$ and $|x-3|=-(x-3)$.

The equation obtained in this case is: $x^{2}-4 x+4+\left(x-1-x+3-\frac{15}{4}\right)^{2}=\frac{65}{16}$. After simplification, we get $x^{2}-2 x+1=0$, from which $x=1$ (but $x=1$ is on the boundary of the interval, so it is a valid solution).

iii) If $x > 3$, then $|x-1|=x-1$ and $|x-3|=x-3$.

The equation obtained in this case is: $x^{2}-7 x+12=0$, from which $x=4$ (and $x=3$ is not in the examined domain).

Our steps are reversible, so the obtained values: $0, 1, 3, 4$ are the solutions of the equation.