Why is finding the roots of a polynomial equation so important? What is to gain?
Maybe the question is: why work on solving something equal to zero, when usually you are interested in solving where you get a value other than zero, or even in solving an equation where neither side is constant?
This is because: it was discovered that equations we are interested in solving can be transformed into equivalent equations where one side is zero. So if we can solve that case, then we can solve other cases, too!
Before the middle ages (in Europe, perhaps earlier in other places) when negative numbers were incorporated into mathematics, there was indeed the need for doing many different types of equations. For the quadratic equation, you would have to learn how to solve these cases: $$ ax^2=bx+c, \ \ ax^2+bx=c, \ \ ax^2+c=bx, \ \ ax^2=bx, \ \ ax^2=c, $$ But now, with the use of negative numbers, we transform all of these (by simple subtraction) to the single case $$ ax^2+bx+c=0 $$ and learn one method to do them all*. (Even before then they knew if there were many terms with constant times $x^2$ to combine them into one term. If there are $x^2$ terms on both sides of the equation, subtract the smaller one. And similarly for $x$ terms and constants.)
*And in the darkness bind them
In a word: factoring. In logical terms: $ab=0\implies a=0$ or $b=0$.
To the extent that algebra concerns itself with setting up and solving polynomial equations, bringing everything to one side, leaving $0$ on the other side, is often an effective way to find all the solutions of an equation -- and, importantly, to know you haven't overlooked any. For example, consider the equation
$$(x^2+11)x=6(x^2+1)$$
If you play around with it, you might notice that $x=1$ is a solution, since $12\cdot1=6\cdot2$. If you play some more, you might find a couple of other solutions. And if you know some theory, you might know when you've found all the solutions. But it becomes systematic, and easier, if you rewrite the equation as
$$x^3-6x^2+11x-6=0$$
The observation that $x=1$ is a solution means that the cubic on the left hand side factors:
$$(x-1)(x^2-5x+6)=0$$
In fact, the cubic in this case factors completely, giving us
$$(x-1)(x-2)(x-3)=0$$
so this tells us that $\{1,2,3\}$ is the complete solution set to the equation $(x^2+11)x=6(x^2+1)$.
Not every equation works out so nicely, of course. But at the very least, setting things up to solve for roots is a systematic way of chipping away at problems involving polynomials.