jmc

We start by completing the square: $$\begin{array}{rl}{x}^2-14x+3& =x^2-14x+{\left(\frac{14}{2}\right)}^2-{\left(\frac{14}{2}\right)}^2+3\\ & =x^2-14x+7^2-49+3\\ & =(x-7{)}^2-46.\end{array}$$Since the square of a real number is at least 0, we have $$(x-7{)}^2\ge 0,$$where $(x-7{)}^2=0$ only if $x=7$. Therefore, $(x-7{)}^2-46$ is minimized when $x=\boxed{7}.$