If we define a polynomial $f(z) = az^2 +bz +c$ (where $z \in \mathbb C$),
then $$f(z) = a(z - p)^2 - \frac{\Delta}{4a},$$
where $p = -\frac{b}{2a}$ and $\Delta = b^2 - 4ac.$

If $a,$ $b,$ and $c$ are all real, then $p$ and $\Delta$ are real.
It follows that $f(x)$ is real if and only if $(z-p)^2$ is real.
This can happen in either of two ways:
$z-p$ is real, or $z - p$ is purely imaginary.

In the case where $z-p$ is real, we can set $z = x$ where $x\in\mathbb R,$
and you can plot $y = f(x)$ over $x\in\mathbb R$ as the question
has done for $a=1,$ $b=-4,$ and $c=13.$
We get a parabola with a vertex at the coordinates
$\left(p, - \frac{\Delta}{4a}\right).$

In the case where $z - p$ is purely imaginary,
we can set $z = p + ix$ where $i$ is the imaginary unit and
$x\in\mathbb R.$
Then
\begin{align}
f(z) &= a(p + ix)^2 + b(p + ix) + c \\
&= ap^2 + i2apx + i^2 ax^2 + bp + ibx + c \\
&= \frac{b^2}{4a} - ibx - ax^2 - \frac{b^2}{2a} + ibx + c \\
&= - ax^2 - \frac{b^2}{4a} + c \\
&= - ax^2 - \frac{\Delta}{4a} \\
\end{align}

If we set $g(x) = - ax^2 - \frac{\Delta}{4a},$
we can plot $y = g(x)$ in a Cartesian plane;
it is a parabola with a vertex at
$\left(0, - \frac{\Delta}{4a}\right),$
oriented in an opposite direction to the parabola we got
for real values of $z.$

So imagine that we want to plot $y = f(z)$ over the complex plane,
but only for the values of $f(z)$ that are purely real.
We can visualize this in just three dimensions:
two dimensions to plot the complex input value, $z,$
and one dimension to plot the real output value $y.$
If we write $z = u + iv$ we can describe this three-dimensional space
by the coordinates $(u,v,y).$

The resulting plot consists of just two parabolas:
the parabola for $y = f(x)$ where $x$ is real,
presented as the plot $y = f(u)$ in the $u,y$ plane;
and the parabola for $y = g(x),$ translated $p$ units in the
$u$ direction and rotated $90$ degrees around the line $u = p,$ $v=0$
(a line through $(p,0,0)$ parallel to the $y$ axis)
so that it lies in the plane $u = p.$

That is, the parabolas have the same vertex,
$\left(p, 0, - \frac{\Delta}{4a}\right),$
and the same axis, the line $u=p,$ $v=0,$
but they are in perpendicular planes.
In another answer to this question,
this same plot is "flattened" into two dimensions by rotating
the "imaginary" parabola by $90$ degrees around its axis.

Now, depending on whether $\frac{\Delta}{4a}$ is positive or negative,
we have the common vertex of the parabolas either below or above the
$u,v$ plane, and we have either two zeros on the upper parabola or
two zeros on the lower parabola.
That translates to two zeros on the line $v=0,y=0$
(two real solutions of $f(z) = 0$)
or two zeros on the line $u=p,y=0$
(two conjugate complex solutions of $f(z) = 0$).
Which of these pairs corresponds to the upper parabola and which to the lower parabola depends on the sign of $a.$
Of course if $\frac{\Delta}{4a} = 0$ then we get the double root $z = p.$