TITLE: Prove that $T_n = 3n^2 -60n + 301$ is positive for every $n$
QUESTION [7 upvotes]: I recently did a Mathematics exam from a previous year, and I stumbled across a question's answer I struggled to fully understand.
It is given: The quadratic pattern $244 ;~ 193 ;~ 148 ;~ 109;~ \ldots$
I've determined the $n$-$\textrm{th}$ term as $T_n = 3n^2 -60n + 301$.
Now the questions asks:

Show that all the terms of the quadratic pattern is positive.

Our teacher explained completing the square of the formula, but I couldn't catch what she said, and rather, I didn't understand why you would complete the square. I do however see that for $3n^2 - 60n + 301$ we have $\Delta < 0.$
I can deduce that you would go in the direction of an inequality whereas you have $n >0$ (incomplete).
Perhaps someone with a higher understanding can explain this to me.

REPLY [4 votes]: A very elementary way to look at this is to compare the similar formula $3n^2 - 60n$. This actually dips down to $-300$ before rising back up to 0 and then it looks like it just keeps climbing up and up.
In order for this to be negative, you need $3n^2 < 60n$. If $n < 60$, then clearly $n^2 < 60n$. But if $n > 60$, then it's obvious that $n^2 > 60n$, so tripling $n^2$ only widens the gulf.
So 301 is the smallest number you can add to $3n^2 - 60n$ in order to bring it up above 0 for the few positive values of $n$ it does go below 0.
As for negative values of $n$, well, $n^2$ is positive anyway and so is $-60n$.