There are two things going on here. First, $\displaystyle \frac{4!}{2^{2}6^{4}}$ is about $0.0046$, not $0.46$.
Second, the problem described by Feller and your simplification to four dice are not the same problem. Feller's problem requires that each face appear twice. Your four-dice problem requires that $some$ two faces appear twice. There's only one way that Feller's event can be satisfied; namely, that each of 1, 2, 3, 4, 5, and 6 appear exactly twice. There are many more ways that some two faces could appear twice. For instance, you could have two 1's and two 2's, two 1's and two 3's, etc. In fact, there are $\binom{6}{2} = 15$ different ways to choose the two faces that will appear.
And that solves your problem. Taking the correct result from Feller, $\displaystyle \frac{4!}{2^{2}6^{4}} \approx 0.0046$, and multiplying it by the $15$ different ways to choose the faces gives $\displaystyle \frac{90}{6^4} \approx 0.0694444$.
With respect to your question about the multinomial coefficient, $\displaystyle \binom{n}{k_1, k_2, \ldots, k_m}$ calculates the number of ways to partition a set of $n$ items into $m$ groups with $k_1$ items in the first group, $k_2$ items in the second group, and so forth. So in Feller's problem $\displaystyle \binom{12}{2, 2, 2, 2, 2, 2}$ is calculating the number of ways to put two dice in the 1's group, two dice in the 2's group, and so forth.
In the four-dice problem you describe, you haven't specified a particular partition of dice into groups; there are several partitions that will satisfy "two faces appearing twice." So, in the four-dice problem, $\displaystyle \binom{4}{2, 2}$ counts the number of ways to put two dice in a 1's group and two dice in a 2's group. It also counts the number of ways to put two dice in a 1's group and two dice in a 3's group, and it counts the number of ways to put two dice in a 1's group and two dice in a 4's group, and so forth. So to calculate the total number of ways of obtaining "two faces appearing twice" you need to multiply $\displaystyle \binom{4}{2, 2}$ by the number of ways to pick two of the faces out of six possible faces, which is $\binom{6}{2}$. Then you mutiply by $\frac{1}{6^4}$ to get the probability you want.
If you want a general answer, the probability of throwing $n$ $d$-sided dice and having exactly $m$ faces appear $\frac{n}{m}$ times would be
$$\binom{n}{m} \frac{n!}{((\frac{n}{m})!)^m d^n} .$$
We show the path to the answers using somewhat tedious listing. The symmetry between $a$ and $c$ could be used to cut down on the work. We use the fact that the quadratic has real solutions if and only if the discriminant $b^2-4ac$ is $\ge 0$, and equal solutions if and only if $b^2-4ac=0$.
Equality is easiest. This happens if and only if $b^2=4ac$. That forces $b=2$, $4$, or $6$. If $b=2$, we need $a=c=1$, so the only configuration is $(1,2,1)$. If $b=4$, we want $ac=4$, which can happen in $3$ ways, $(1,4,4)$, $(4,4,1)$, and $(2,4,2)$. Finally, if $b=6$, we want $ac=9$, which only happens with the configuration $(3,6,3)$. Each configuration has probability $\frac{1}{6^3}$, so the required probability is $\frac{5}{216}$.
For real solutions , we want $b^2\ge 4ac$. That cannot happen if $b=1$. If $b=2$, it can only happen if $ac=1$, giving a contribution of $\frac{1}{216}$. If $b=3$, we want $ac\le 2$, which can happen in $3$ ways, for a contribution of $\frac{3}{216}$.
We leave the cases $b=4$ and $b=5$ to you. For $b=6$, we want $ac\le 9$. Let us list the ways. With $a=1$, $b$ can have $6$ values. With $a=2$ there are $4$. With $a=3$ there are $3$. With $a=4$ there are $2$. And there are $1$ each for $a=7$ and $a=6$. That gives a contribution of $\frac{17}{216}$.
For complex, one could say that the probability is $1$, since every real number is in particular a complex number. But what is probably intended is complex and non-real. Then the required probability is $1$ minus the probability the root(s) are real.
Best Answer
The formula for the roots is $$ x = \frac{ -b \pm\sqrt{b^2 - 4ac}}{2a}. $$ The real/complex dependence of the roots of the coefficients therefore only depends on $b^2 - 4ac$. Now since $1 \le a,b,c \le 6$ and that those are integers, there are not many possibilities for this to equal zero. By inspection,
$1 - 4ac \ge 0$ means $\frac 14 \ge ac$, which is impossible, hence we have $36$ complex-root-cases
$4 - 4ac \ge 0$ means $1 \ge ac$, which means $a=c=1$, hence $(1,2,1)$ is one equal-root-case (and real), or the other $35$ cases are complex
$9 - 4ac \ge 0$ means $\frac 94 \ge ac$, so that equality is impossible, but we can have $ac = 1$ and $ac=2$ to satisfy this inequality, hence there are three real-roots-cases here ($(1,3,1)$, $(1,3,2)$, $(2,3,1)$) and $33$ other complex-roots-cases
$16- 4ac \ge 0$ means $ac \le 4$, hence $(4,4,1)$, $(2,4,2)$ and $(1,2,4)$ are 3 equal-root-cases, the number of ways to get $ac \le 4$ (if you count them, you get 11, 12, 13, 14, 21, 22, 31, 41, so there are $8$) is the number of real-root-cases, and the rest ($28$ cases) are complex
$25- 4ac \ge 0$ means $\frac {25}4 \ge ac$, thus equality is impossible, but the number of ways to get $ac \le 6$ is $14$ (11,12,13,14,15,16, 21, 22, 23, 31, 32, 41, 51, 61 are the $14$ cases) hence the $22$ other cases are complex-roots-cases and those $14$ are real-root-cases
$36- 4ac \ge 0$ means $ac \le 9$, hence $(3,6,3)$ is one equal-root-case and the real root cases are those with $ac \le 9$, hence giving the $15$ cases as for $ac \le 6$, plus the 3 cases 24, 33 and 42, for a total of $17$ real root-cases and $19$ complex-root-cases. To sum it up,
Therefore the probability that the roots are equal is $\frac{5}{6^3} = \frac 5{216}$, the probability that the roots are real is $\frac{1+3+8+14+17}{6^3} = \frac{43}{216}$ and the probability that the roots are complex is $\frac{36+35+33+28+22+19}{6^3} = \frac{173}{216}$.
Hope that helps,