Prove that $T_n = 3n^2 -60n + 301$ is positive for every $n$ 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. 

First, we know all squares are positive.
$$(n-10)^2 \ge 0\\
3(n-10)^2 \ge 0 \\
3(n-10)^2 +1 \ge 1 >0 \\
3(n-10)^2 +1  >0 $$
By expanding we will get
$$3n^2-60n+301>0$$
The way you should approach this is by thinking: what must be positive? squares. Then, how do I come up with a square? Completing the Square.
Completing the Square is explained well here.
