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. 
 A: Completing the square gives you $ \Delta \leq 0$
It shows that equation will always be positive.  
But, it is easier to do  so using calculus.
Take $$y=3n^2-60n+301$$
Take first derivative with respect to n: $$\frac{dy}{dn}=y'=6n-60$$
$$y'=6(n-10)$$
As you can see, for all $$n <10 ,  y'<   0$$ 
And, for all $$n>10, y'>0$$ 
It means that function $y=f(x)$ attains its minimum at x=10. f(10)=1
It decreases before $x=10$ and increases after $x=10$.  
So, it can never be negative.    
have a look at it's graph
A: 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 $-60n$.
A: Completing the square enable you to see the vertex clearly. For $f(x)=ax^2+bx+c$ where $a>0$, we can see the minimal value that it can attain.
Alternatively, just see that the discriminant is negative, that is the function doesn't intersect the $x$-axis and it doesn't change sign. Since one of the term is positive, every term is positive.
A: 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.
