Dual plot for complex roots of quadratic equation Real roots of quadratic equation
$ x^2 - \sqrt 3 x +  1/2 =0  \tag{1} $
can be plotted on  $x$- axis  as its  parabola intersection at $ (\sqrt 3/2 \pm 1/2,0). $
In an improvization I assign $x$-axis as real $x$-axis and $y$-axis as imaginary axis. For plotting complex roots it is possible to plot roots of quadratic equations on same $x$- but different $y$- as a  dual representation (convenient) mode converting the Plot to an Argand/Gauss diagram ?
That is, is there some geometric way to  represent complex roots  $( \sqrt 3 /2, \pm\, i/2)  $ of
$$  x^2 -\sqrt 3 x  +1  =0 $$   as well on such dual diagram?
What geometric constructions may be necessary to make any new line cut the parabola or its morph at the complex roots? The situation shown:


I am not sure if I understood well your question. So forgive me if the following answer does not satisfy you.
Anyway, if you have an equation like $$x^2-6x+10=0$$ not having real solutions, you can still plot the function $$f(x)=x^2-6x+10.$$ See the blue curve below. (The coefficient belonging to $x^2$ will be $1$ during this argumentation.) The fact that the original equation does not have real solutions is equivalent to the fact that $f$ does not intersect with the $x$ axis. Mirror the plot of $f$ over the line tangent to its minimum being at $m=3$. The resulting (red) curve will have intersections with the $x$ axis (at $3\pm 1$ or $m \pm 1$ this time).

The complex solutions of the original equation are $3\pm i.$ (Where the multiplier of $i$ is $1$. Note that the intersection points are at $m\pm 1$.)
This is true in general: Suppose that $f(x)=x^2+bx+c$ does not intersect the $x$ axis and that its minimum is taken at $m$.  Mirror $f$ over its tangent line at $m$. The resulting function $g(x)=2f(m)-f(x)$ will intersect the $x$ axis at points, say $m+u$ and $m-u$. The complex solutions of $x^2+bx+c=0$ are then
$$m \pm iu.$$
That is, the complex roots can be constructed without having to use the solution formula.
Consider the equation given in the OP:
$$  x^2 -\sqrt 3 x  +1  =0.$$
Here the blue line represents the function $x^2 -\sqrt 3 x  +1  =0$ whose minimum is at; $m=\frac{\sqrt{3}}{2}.$

The purple line is the mirror image whose intersection points with the $x$ axis are at $\frac{\sqrt{3}}{2}\pm\frac{1}{2}$ so the complex roots are
$$\frac{\sqrt{3}}{2}\pm i\frac{1}{2}.$$
