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,
- $b=1$ : $0$ equal, $0$ real, $36$ complex
- $b=2$ : $1$ equal, $1$ real, $35$ complex
- $b=3$ : $0$ equal, $3$ real, $33$ complex
- $b=4$ : $3$ equal, $8$ real, $28$ complex
- $b=5$ : $0$ equal, $14$ real, $22$ complex
- $b=6$ : $1$ equal, $17$ real, $19$ complex.
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 $\frac{36+35+33+28+22+19}{6^3} = \frac{173}{216}$.
Best Answer
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.