Why are zeros/roots (real) solutions to an equation of an n-degree polynomial? I can't really put a

I'll try to interpret your question from the data Peter managed to extrapolate from you. It seems to me that your confusion stems from the difference between a variable and an unknown, as it is mentioned in one of the comments. Why we are interested in precisely the x at which the $f(x)$ evaulates to 0 should become clear to you as you read through this answer. I apologize if I oversimplify it, I cannot infer your knowledge.
Linear interpretationLet's go down a degree from your examples, to a simple linear equation where x (the unknown) is related to some value b by a factor a where $a\ne 0$. Expressing this as an equation is straightforward:
ax=b
We wish to know the value of x, for which the equation is valid (true) or, put a little bit different, x multiplied by a is equal to b. This can be easily resolved by simple algebra and logical thinking because b is a times bigger or smaller than x (remember that a can be anything except 0, that includes numbers above 0 and below 1 which would indicate that b is smaller than x because it is only a fraction of its value). Therefore, if we divide both sides by an equal amount (a), we can find out what x is:
$x=\frac{b}{a}$
And there, we have the fancy x. But this is where stuff gets fun. I believe you are acquainted with the notion of functions, which take a certain input in a certain range and return an output which is calculated from that input, thus forming an ordered pair which can be graphed. The magic is that the output (dependent value) depends on the input (independent value) and we can study their relationship visually or purely algebraically.
In our former relationship, we have two coefficients/constraints (known values which force our x (the unknown) into a very particular value (the solution to our problem, an answer to our question, what is x?). You can clearly see that it can be interpreted in this form:
$f(x)=ax$
$f(x)=b$
Now, we want to know at which x will the function f(x) output the value b. Now, if may turn your attention to this, b is also a function of x. That's right, for every x you can think, it simply pops out b. It is constant all across all x values. Well, how can we achieve that? Take the x, drop the x:
$g(x)=b$
You know b, so for any x, it just outputs b, which you can imagine in your head or on a piece of paper as a simple line that is parallel to the x-axis. Now graph out the function f(x)=ax on that mental or paper image. You will notice that they intersect at a very specific point. At that point, they have the same x and y values. Since we know that g(x)=b, their shared y coordinate is b. That's the whole point of g(x), we wanted to know the x value where the two lines collide at y=b. So, if at that magic point they share the x and y coordinate, a common point then as you know:
$f(x)=g(x)$
So, by making such a statement, we infer that there ought to be an x for which these two functions would yield the same value. Although we suppose there exists such a value, we can detect that they are parallel (that there is no solution). A proof by contradiction, if you will. Substituting, we find out that:
$ax=b$
$x=\frac{b}{a}$
Which is exactly what we reached with simple algebra. So, graphically, the intersection point is the solution to this equation. They are the same, g(x) and f(x) are the same at that specific x if and only if:
$f(x)-<!--\; -\; -->g(x)=0$
or
$ax-<!--\; -\; -->b=0$
That's a visual way to see where the 0 comes from. If the two functions are the same or EQUAL, subtracting one from the other MUST yield 0, which is only logical. From simply observing the equation ax=b, they are only equal if subtracting one from the other yields 0. Basically, what happens really is that you subtract from both sides one of the sides. By doing the same operations on both sides, we don't damage the equation. In everyday life, we simplify by saying, subtract one side from the other, going over the equal sign like a country border, changing sign as a passport.
Going up a degree...Now, let's expand this to quadratic equations and thus functions. As you know, a quadratic equation is defined implicitly as:
$a{x}^2+bx+c=0$
I hope by now you're beginning to notice the difference between unknowns and variables. Variables give away their nature with their name, they can change and evaluate to different values which all respect the relationship imposed by a function. Since a function works with the full range of values x can take, x is a variable. And with x, the independent value, there comes a value which is a function of that x, f(x) or y.
The moment you impose a specific x (input) or a specific f(x)=y (output), the other one can take only one very specific value which forms an ordered pair with it. And the other one is an unknown, which can be calculated by the relationship described in the equation (constants/coefficients/constraints force the value).
Without beating a dead horse (too much at least), consider a quadratic function whose graph is a parabola. Simply by using the already mentioned here (and understanding the shape of the parabola), you can infer why there are two solutions. The g(x) or the imposed y is still a horizontal line. And that line cuts the parabola in two points, both equally valid. A simple argument for formulizing the quadratic equation is to solve problems like the value of our dear $\sqrt{2}$ which is an irrational number, which means it's impossible to represent it as a ratio of two values. But luckily, it is not transcendental, so it can be expressed as:
$x=\sqrt{2}$
$x^2=2$
$x^2-<!--\; -\; -->2=0$
Remember, we are trying to find an x which when squared gives the value of 2. Now, what is this 2? It is a constraint which forces x to evaluate to $\sqrt{2}$. Rings a bell? Yes, it's the same form as f(x)−g(x)=0. The intersection. Or multiple ones in this case (two points/ordered pairs sharing the same y value), as you've already seen visually on your paper. Now, let's get a bit dirty by deriving the actual equation which releases the x value to us.
$a{x}^2+bx+c=0$
$x^2+\frac{b}{a}x=-<!--\; -\; -->\frac{c}{a}$
We can complete the square by adding to both sides $\frac{b^2}{4a^2}$ (hopefully you can see that):
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-<!--\; -\; -->\frac{c}{a}+\frac{b^2}{4a^2}$
$(x+\frac{b}{2a}{)}^2=\frac{b^2-<!--\; -\; -->4ac}{4a^2}$
Slowly shaping up to the classic equation, just to take square roots from both sides:
$x+\frac{b}{2a}=\frac{\sqrt{b^2-<!--\; -\; -->4ac}}{2a}$
$x=\frac{-<!--\; -\; -->b+\sqrt{b^2-<!--\; -\; -->4ac}}{2a}$
HOWEVER!
Remember how we decided it would be best if two negative numbers multiplied yield a positive number, because that's what logic and everyday problems dictate. Well, now we must take one for the team. The square root of a number is a number multiplied by itself. Therefore, that number can be either positive or negative. Since we have no way of knowing, we must then have two distinct values and therefore two perfectly valid solutions which can only be culled by further application of logic (if it's a simple case, for example length, well, it can't be negative). But that's not always the case, therefore, we leave this:
$x_1=\frac{-<!--\; -\; -->b+\sqrt{b^2-<!--\; -\; -->4ac}}{2a}$
$x_2=\frac{-<!--\; -\; -->b-<!--\; -\; -->\sqrt{b^2-<!--\; -\; -->4ac}}{2a}$
And that's the mystery of two solutions resolved from a more algebraic-consequence point of view. This is the most general case, you can sometimes be without the y-intersect (c coeffiecient) or the b coefficient like in the case of $x^2-<!--\; -\; -->2=0$ which you can now evaluate to be somewhere around:
$\sqrt{2}=1.4142135623730950488016887242097\dots <!--\; \dots \; -->$
So, that should answer your question of why 0? or the x-intercept, why do they correspond to the solutions and now for the final one:
$a{x}^2+bx+c=403045$
After all that has been discussed, this is obviously a totally different equation. You're now evaluating a totally different ordered pair, aren't you? I hope you see that you're looking for the x (or xs) that correspond to the y=403045. That's a totally different intersect. You've already said that something is equal to 0, that they intersect and you're overdoing it with a second uplift of the value.
If you're looking for a value that is described by some function $f(x)$, this is entirely valid and has a solution of its own which yields true to the equation (the equal sign, heart of everything). But that has nothing to do with solving a specific problem, that's just trying to evaluate the function in the other way around, by finding x by the it's y value (the output).
Hopefully this helps! Unfortunately, it's a bit long. Need to work on that.