Find the minimum value of $x^2-6x+10$ for real $x$.

**Complete the square:** $x^2-6x+10=(x-3)^2+1$. Since $(x-3)^2\ge 0$ for all real $x$, the minimum value occurs when $(x-3)^2=0$, that is at $x=3$. **Therefore the minimum value is $1$.**