The quadratic case can be dealt with as follows. A quadratic polynomial $f(x) = ax^2 + bx + c \in \mathbb{Z}[x]$ has two distinct real roots if and only if $\Delta(f) = b^2 - 4ac > 0$, and a pair of complex conjugate roots if and only if $\Delta(f) < 0$.
We now let $a,b,c$ vary in the box $[-X,X]^3$. We first pick a pair $(a,c) \in \mathbb{Z}^2 \cap [-X,X]^2$. If $ac < 0$, that is, if the pair $(a,c)$ lies in two of four quadrants, then $\Delta(f) < 0$; hence 100% of quadratic polynomials with $a,c$ coming from those two quadrants have two real roots. The remaining two quadrants are symmetric to each other, so we might as well consider only the positive quadrant. We can exploit symmetry once more to assume that $a \leq c$, and from density considerations we can carve out the 0-density sets corresponding to $a = 0$ and $a =c$; whence we assume $0 < a < c$.
The count of triples $(a,b,c)$ satisfying $0 < a < c \leq X$ and $\Delta(f) < 0$ can be estimated by the triple integral
$$\displaystyle \int_1^X \int_1^c \int_{-2\sqrt{ac}}^{2 \sqrt{ac}} db da dc = \frac{8}{9} X^3 + O(X^2).$$
Multiplying by 4 to account for the assumption that $a \leq c$ and $a,c > 0$ (and using standard geometry of numbers arguments), we see that the total number of negative discriminant quadratic polynomials of height at most $X$ is
$$\displaystyle N^+(X) = \frac{32}{9} X^3 + O(X^2).$$
The number of positive discriminant forms is then
$$\displaystyle N^{-}(X) = 8X^3 - \frac{32}{9} X^3 + O(X^2) = \frac{40}{9} X^3 + O(X^2).$$
One can do a similar (but much more difficult) argument for cubic polynomials (binary forms), by exploiting the fact that for a cubic binary form $g(x,y) \in \mathbb{Z}[x,y]$, its Hessian covariant $q_g(x,y)$ (which is a quadratic form) has discriminant $-3\Delta(g)$; and hence the problem of counting cubic binary forms with three or one real linear factors is reduced to dealing with the Hessians. However, the inequalities involved are no longer linear in general, and hence the application of geometry of numbers methods will be more complicated. Cremona also worked out the exact conditions for quartic polynomials to have 0, 2, or 4 real roots in https://homepages.warwick.ac.uk/~masgaj/papers/r34jcm.pdf
I suspect the methods I used above become intractable very quickly with respect to the degree, so perhaps a different formulation is necessary to make progress.