Solve the following equation in the set of real numbers:

$$
\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}
$$
In order for the expressions in the equation to be meaningful, the following conditions are necessary:

$$
a, b, c, a-b+c \geq 0
$$

After these, transform the equation. Squaring

$$
a-b+c=a+b+c+2 \sqrt{a c}-2 \sqrt{a b}-2 \sqrt{b c}
$$

rearranging

$$
0=b+\sqrt{a c}-\sqrt{a b}-\sqrt{b c}
$$

Factoring

$$
0=(\sqrt{b}-\sqrt{a})(\sqrt{b}-\sqrt{c})
$$

The value of a product is exactly 0 if and only if one of its factors is 0, that is,

$$
\sqrt{b}=\sqrt{a} \text { or } \sqrt{b}=\sqrt{c}
$$

thus $b=a$ or $b=c$.

In our steps, we did not lose any solutions, but our transformations were not necessarily equivalent; thus, the result we obtained must be verified. However, it is clear that under our initial assumptions, whether $b=a$ or $b=c$, the equation holds (the condition $a-b+c \geq 0$ follows from both $b=a$ and $b=c$). Therefore, the solutions are:

$$
a=b \geq 0, c \geq 0 \text { or } a \geq 0, b=c \geq 0
$$

Based on the work of Attila Törös (Tata, Eötvös J. Gymn., II. o. t.)