3. Solve the equation:

$$
\sqrt{x-3}+\sqrt{7-x}=x^{2}-10 x+23
$$
3. Let's make the substitutions $t_{1}=\sqrt{x-3}$ and $t_{2}=\sqrt{7-x}$. Clearly, $t_{1}, t_{2}>0$. Note that $(x-3)(7-x)=-x^{2}+10 x-21$, i.e., the right side of the given equation can be written as $-t_{1}^{2} t_{2}^{2}+2$; thus, we have $t_{1}+t_{2}=-t_{1}^{2} t_{2}^{2}+2 \leqslant 2$. On the other hand, since $t_{1}^{2}+t_{2}^{2}=x-3+7-x=4$, we have $\left(t_{1}+t_{2}\right)^{2}=t_{1}^{2}+2 t_{1} t_{2}+t_{2}^{2}=4+2 t_{1} t_{2} \geqslant 4$, from which we get $t_{1}+t_{2} \geqslant 2$. From the two obtained equalities, it follows that $t_{1}+t_{2}=2$, and we see that this is achieved only in the case $t_{1} t_{2}=0$, i.e., $t_{1}=0$ or $t_{2}=0$. From this, we obtain that the only possible solutions to the original equation are $x=3$ and $x=7$, and by verification, we confirm that both of these numbers are indeed solutions.