On discriminants and nature of an equation's roots? Edited: All equations in the post are assumed to have all real coefficients and are minimal polynomials.
While trying to ascertain if the Brioschi quintic $B(x)=x^5-10cx^3+45c^2x-c^2=0$ could ever have $3$ real roots, I was led to the question if one can use the discriminant $D$ to settle this. 
For $B(x)$, it is given by $D = 5^5c^8(-1+1728c)^2$ and it seems for quintics, if $D>0$, then there are either $0$ or $4$ complex roots $C=a+bi$ with $b\neq0$. Hence the Brioschi (with real coefficients) can never have $3$ real roots.
For other degrees $n$, by observing the data in the Database of Number Fields, I was able to come up with the table below. The second and third columns give the number of complex roots $C=a+bi$.
$$\begin{array}{|c|c|c|}
\hline
\text{Degree}\;n&\text{If}\;D>0&\text{If}\;D<0\\
2&0&2\\
3&2&0\\
4&{0,4}&{2}\\
5&{0,4}&{2}\\
6&2,6&0,4\\
7&0,4&2,6\\
8&{0,4,8}&{2,6}\\
9&{0,4,8}&{2,6}\\
{10}&2,6,10&0,4,8\\
{11}&0,4,8&2,6,10\\
{12}&{0,4,8,12}&{2,6,10}\\
{13}&{0,4,8,12}&{2,6,10}\\
{14}&{0,4,8,12}&{2,6,10,14}\\
{15}&{0,4,8,12}&{2,6,10,14}\\
\hline
\end{array}$$
Questions:


*

*Is the table true? 

*How do we predict the second and third columns for much higher $n$? For example, for $n=163$, does the second column start as $0,4,8,12,\dots$ or $2,6,10,14,\dots$?

 A: I believe that there is a mistake in your table.
Brill's theorem states that the sign of the discriminant of an algebraic number field is $(-1)^{r_2}$ where $r_2$ is the number of complex places. When we have a power basis for our number field, the minimal polynomial of the generator will have $2r_2$ complex roots. Thus the column for $D>0$ should only contain integers divisible by $4$, and when $D<0$ we have a number of complex roots $\equiv 2\pmod{4}$.
For a specific example, $$x^3-x^2-3x+1$$ has three real roots and a discriminant of 148.
A: I don't think the table is correct. $f = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$ has discriminant $1194393600 > 0$ and 0 complex roots.
It doesn't have a direct citation from a reputable source, but the Wikipedia page you linked says that in general for $D > 0$ there is an integer $0 \leq k \leq \frac{n}{4}$ such that there are $4 k$ complex roots, and for $D < 0$ there is an integer $0 \leq k' \leq \frac{n-2}{4}$ such that there are $4k' + 2$ complex roots, which doesn't mesh with your table.
