Determine the form of set containing the real solutions 
From the following inequality:
  $\dfrac{\log_{2^{x^2+2x+1}-1}(\log_{2x^2 + 2x + 3}(x^2 - 2x)}{\log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)} \geq 0 $, the set of all real solutions to this inequality is of the form:
  $$(a) \ (a,b) \cup (b,c) \\ (b) \ (-\infty,a) \cup (c,\infty) \\ (c) \ (a,b) $$
  for some real numbers $a,b,c $ such that $-\infty<a<b<c<+\infty $.

First I found the points at which the numerator and the denominator equal $0$. I found that from the following:
$$\log_{2^{x^2+2x+1}-1}(\log_{2x^2 + 2x + 3}(x^2 - 2x) = 0 \\ \log_{2x^2 + 2x + 3}(x^2 - 2x) = 1 $$
$$2x^2 + 2x + 3 = x^2 - 2x .$$
From this quadratic equation i find that the zeros of the function are $x_1=-3 $ and $x_2=-1$ and let them be $a=-1 $ and $b=-3 $ for now. I also found from the plot of the given quadratic equation that $(-\infty,-3)\cup(-1,\infty) $ makes function a positive one and the values in between $(-3,-1)$ make it negative. I used that later for the rational number chart.
Similarly $$\log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10) = 0 \\
x^2 + 6x + 10 = 1 \\ x = -3.$$ This function is positive for every other value.
Now using the chart of rational numbers, determining when the whole ratio is positive and when is negative I get the positive values to range from:
$(-\infty,-3) \cup (-1,\infty)$ which then makes my answer of the form $(-\infty,b) \cup (a, +\infty)$.
First of all, it's not given, second, it's wrong. The solution I have is for the $(a)$ answer. I just don't know how to get to it. 

Do you know about the change of base formula? If $a,b,x > 0$, $a \not= 1$, and $x \not= 1$ then
$$ \log_a b = \frac{\log_x b}{\log_x a}.$$ Your original inequality reads
$$ \log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0$$ provided that $2^{x^2 + 2x + 1} - 1 > 0$, $2^{x^2 + 2x + 1} - 1 \not= 1$, $x^2 + 6x + 10 > 0$, $x^2 + 6x + 10 \not= 1$, $\log_{2x^2 + 2x + 3} (x^2 - 2x) > 0$, and $2x^2 + 2x + 3 > 0$.  These are all true provided that $x \not= 0$, $x \not= -2$, $x \not= -1$, $x \not= -3$, and $x^2 - 2x > 1$.  
Next observe $$ \log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0 \iff \log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1 \\ \iff x^2 - 2x \ge (2x^2 + 2x + 3)^1.$$
The last inequality reduces to $x^2 + 4x + 3 \le 0$ which has the solution $[-3,-1]$. Note that $x^2 - 2x \ge 3$ on this interval, so excluding $x = -1$, $x = -2$,  and $x = -3$ you get $$ (-3,-2) \cup (-2,-1).$$
