How do you complete the square for $2x-x^2$?
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In case you want the general method,
Start with $$ax^2+bx+c,\qquad a\ne0$$
Factor out the $a$ (which may be negative, as in $2x-x^2$) from the first two terms: $$a(x^2+(b/a)x)+c$$
Add and subtract the square of half the coefficient of $x$: $$a(x^2+(b/a)x+(b/(2a))^2-(b/(2a))^2)+c=a(x^2+(b/a)x+(b/(2a))^2)+c-(b^2/(4a))$$
Factor: $$a(x+(b/(2a)))^2+c-(b^2/(4a))$$ Voila! The square, she is completed. When you've had as much experience as Ross and The Chaz, you'll be able to just look at a quadratic and write down the completion, but, until then, there's the utter and complete procedure.