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

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.

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 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 $-60n$.