Hello again PG,
Well you assumed that we can only solve for y=0 but actually this kind of method is much more general method than that. You only need to see a few more examples to see how versatile this really is. Of course i have an example for you
We start with the quadratic:
y=ax^2+bx+c
where a, b, and c are constants and i did not show some multiplication signs for simplicity in appearance.
To find out the value of x when y is zero we substitute y=0 and we get:
0=ax^2+bx+c
and when we solve for x we get two values for x where y is zero.
But what is to stop us from using any other value of y? For your second thought we might make y=2 as you said, then we would have:
2=ax^2+bx+c
And now we have four constants, but we can reduce this again to three by allowing 'c' to absorb the '2' as simply as subtracting 2 from both sides:
0=ax^2+bx+(c-2)
And now we have again a quadratic, except now when we solve for x we get the values that cause y to be 2.
But the versatility does not stop there. Suppose we dont actually have a constant value of y, but instead we have a simple linear straight line that angles from the horizontal:
y=D*x
That's a sloping line. Now say we want the solution to x when y meets that line. We start again wth:
y=ax^2+bx+c
but now we also have:
y=Dx
so we can make the subts again into the starting quadratic:
Dx=ax^2+bx+c
and now we allow the bx to absorb the Dx (we just subtract Dx from both sides):
0=ax^2+(b-D)x+c
Now this is just a quadratic again, and when we solve it this time we end up with the solutions for the intersection of the parabola and the straight sloping line.
There are a few catches, for example if we try to solve for y=-2 instead of +2 for a parabola that never goes below zero, we will end up with no real solution.
Of course we can also use two quadratics and ask the question if they intersect:
y=ax^2+bx+c
y=Ax^2+Bx+C
and after equating the two we get:
0=(a-A)x^2+(b-B)x+(c-C)
and the solutions (if any) will be the intersections of the two parabolas.
The quadratic is limited only in the order of the equation, that it has to be 2. Thus if we have the two:
y=ax^2+bx+c
y=Ax^3+Bx^2+Cx+D
this no longer admits a quadratic solution, because of the third degree power, so this becomes a third degree problem where we cant use the quadratic formula (directly) anymore. There are ways to solve the third degree too though using methods for third degree equations.
As to the practical applications for this kind of thing, they are endless in number.
From all your questions you've asked here on ET it appears that you could benefit greatly from a short class (or self study) in precalculus before or during a study in calculus, and a little analytic geometry wouldnt hurt either. You seem to be doing quite well with your calculus though (and really great with your English BTW) but you might find it easier to get through with these few prerequisites.