For which real numbers $x, y, z$ does the equality

$$
x^{2}+y^{2}+\frac{3}{4}=z+\sqrt{x+y-z}
$$

hold?
In order for the right side to make sense, we must ensure that $x+y-z \geq 0$. Let $p=\sqrt{x+y-z}$; then substituting $\left(x+y-p^{2}\right)$ for $z$, the equation transforms as follows:

$$
x^{2}+y^{2}+\frac{3}{4}=x+y-p^{2}+p
$$

Moving all terms to the left side and transforming into a sum of complete squares:

$$
\begin{gathered}
x^{2}-x+y^{2}-y+p^{2}-p+\frac{3}{4}=0 \\
\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}+\left(p-\frac{1}{2}\right)^{2}=0
\end{gathered}
$$

The sum of three non-negative real numbers is exactly 0 if and only if each term is 0, that is,

$$
x=\frac{1}{2}, \quad y=\frac{1}{2} \quad \text { and } \quad p=\frac{1}{2}
$$

and thus

$$
z=x+y-p^{2}=\frac{3}{4}
$$