Now, my doubt is : can the zeroes of any polynomial be found using such forms (as given above)? Maybe some $\frac{-b}{na} (n = \text{degree of the polynomial})$ can be used to deduce the zeroes faster ?
Kind of, yes!
As you know, the quadratic formula can be used to find the roots of any quadratic polynomial. And as you've noticed, the quadratic formula (as you've written it) consists of the term $\frac{-b}{na}$, where $n$ is the degree of the polynomial (which is 2), plus some variable stuff. (By "variable stuff," I mean the part that looks like $\pm \sqrt{\text{something}}$, where choosing a plus sign gives you one root and choosing a minus sign gives you the other one.)
Does the pattern continue? Yes, it does! There is also such a thing as the cubic formula, which can be used to find the roots of any polynomial of degree 3. The cubic formula is somewhat complicated, but it does in fact consist of the term $\frac{-b}{3a}$ plus some "variable stuff," just like the quadratic formula.
After that, there's the quartic formula, which is even more complicated, but which can be used to find the roots of almost any polynomial of degree 4. Just as you would expect, the quartic formula consists of $\frac{-b}{4a}$ plus some "variable stuff".
Unfortunately, there's no formula which can be used for finding the roots of any polynomial of degree 5—at least, there isn't one that uses only the "ordinary" kinds of functions that are used.
Nevertheless, the pattern does continue. The average of all of the roots of a polynomial is always $\frac{-b}{na}$, where $n$ is the degree of the polynomial, $a$ is the coefficient of $x^n$, and $b$ is the coefficient of $x^{n-1}$. (I actually didn't know this before—you've taught me something!)
Here is the proof. Any polynomial can be written as $a(x - r_1)(x - r_2)\cdots(x - r_n)$, where $n$ is the degree of the polynomial. If we multiply this out, it's possible to see that $b$ (that is, the coefficient of $x^{n-1}$) is $-a$ times the sum of all of the roots. As an equation:
$$b = -a \cdot (\text{sum of roots}).$$
By dividing both sides by $-na$, we see that
$$\frac{-b}{na} = \frac{\text{sum of roots}}{n}.$$
The right-hand side is, of course, the average of all of the roots.
