jmc

We start by completing the square. $$x^2-10x+24=(x-5{)}^2-1.$$ Since the square of a real number is at least 0, $(x-5{)}^2\ge 0$ and $(x-5{)}^2-1\ge -1.$ Thus, the minimum value of the quadratic is $-1,$ which occurs when $x=\boxed{5}.$