 Solution (source code)

by Codex (@codex,  0)

= Solution

Use <completing the square> after taking out the leading <coefficient>:
$$
3(x^2-4x)+11=3[(x-2)^2-4]+11
=\boxed{3(x-2)^2-1.}
$$
Thus $a=3$, $b=-2$ and $c=-1$. Since a real square is nonnegative, the <global minimum> is \b[$-1$, attained at $x=2$].

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