Why are zeros/roots (real) solutions to an equation of an n-degree polynomial? I can't really put a proper title on this one, but I seem to be missing one crucial point. Why do roots of a function like $f(x) = ax^2 + bx + c$ provide the solutions when $f(x) = 0$. What does that $ y = 0$ mean for the solutions, the intercept at the $x$ axis? Why aren't the solutions at $f(x) = 403045$ or some other arbitrary $n$?
What makes the x-intercept special?

I believe this is another way of looking at it.

In most cases, "roots" are a subset of all possible solutions (all the $x$, $y$ coordinates that satisfy the equation). If an equation does not have a root solution, then there is no direct, algebraic way of solving for $x$, given a value $y$; but you can still solve for any $y$ (the function's output), given a value $x$.

Consider:
$y = x^2 - 1$;  the root is $1$ when $y=0$. however, $y = x^2 + 1$;  there is no real number algebraic solution when $y=0$. You can not solve for $x$ when the parabola does not intersect the $x$-axis. That's why setting the equation to $x$ is useful to indicate that situation. However, the equation still has an infinite number of solution coordinates by solving for $y$ given a value of $x$. You just can't go the other way around (solve for $x$) given $y$.  It's comparable to trying to solve for $x$ in the equation $y = 5$; but you can solve for $y$, given any value of $x$.  
