jmc

We start by completing the square: $$\begin{array}{rl}2x^2-12x+3& =2(x^2-6x)+3\\ & =2(x^2-6x+(6/2{)}^2-(6/2{)}^2)+3\\ & =2((x-3{)}^2-3^2)+3\\ & =2(x-3{)}^2-2\cdot 3^2+3\\ & =2(x-3{)}^2-15.\end{array}$$ Since the square of a real number is at least 0, we have $(x-3{)}^2\ge 0$, where $(x-3{)}^2=0$ only if $x=3$. Therefore, $2(x-3{)}^2-15$ is minimized when $x=\boxed{3}.$