Roots of a second degree equation.

Roots of a quadratic equation.
y=ax²+bx+c
If b² >4ac then there are two roots.
If b=4ac then there is a single root.
If b²<4ac there are no roots.
We can then draw as many curves as we want, as long as we want, nothing will change, there are no roots.
At least in the real world.
Let us set y=x²+1 or y=x²+4x+5; there are no roots.
This does not exist, looking for roots in nothingness, or rabbit horns will not change anything. We will not find any.
Some mathematicians will then try to find some anyway, but beyond reality, where Doctor Hachel takes possession of your computer screen and will give it back to you only if he wants to (I put a virus in the equation mentioned above).
They call these imaginary roots, because, since they do not exist, we must imagine them. But what do they correspond to?
It would seem, in fact, that they are not the roots of the equation, that is to say the place in x where the curves cross y=0 since it is impossible for all x, but the horizontal mirror projection passing through the vertex of the curve.
For y=x²+1 then [-b(+-)sqrt(b²-4ac)]/2a --->
(+-)sqrt(-4ac)/2
(+-)sqrt(4i²)/2=(+-)i
The two roots are x'=-i and x'=i (i.e. x=-1 and x=1 of the imaginary inverted curve).
Let us set =x²+4x+5 which has no root, and project this curve in mirror; two roots appear for the mirror curve.
[-b(+-)sqrt(b²-4ac)]/2a --->[-4(+-)sqrt(16-4*5)]/2 --->
[-4(+-)sqrt(4i²)]/2a--->(-4(+-)2i)/2
x=-2-i
x'=-3
x"=-1
But this is very interesting, but wouldn't it be worth giving right away the equation of the mirror curve whose vertex touches the vertex of the real curve by specifying that it is the imaginary horizontal mirror?
This means that a second degree curve has two roots (or a single one) and that when it doesn't have any, what crosses y=0 is its imaginary mirror curve which will give two imaginary roots.
R.H.