<s>[INST] How to find the real values of the parameter a, so that the inequality doesn't have positive solutions? The inequality is the following:
$$
\frac{x^2 - 6ax + 2x - 5a - 1}{x+a+1} < 0
$$
If we write this inequality like $\frac{a}{b} < 0$, I may say that for $a=0$, I found that 
$X_1=3a - 1 - \sqrt{9a^2-a+2}$
$X_2=3a - 1 + \sqrt{9a^2-a+2}$
What should I do next? The equation should have only negative solutions.
 [/INST] For a fraction to be negative we need both the numerator and the denominator to have the opposite sign. For example, $(-2)/(+4) = -1/2$ and $(+3)/(-4) = -3/4$.
You need to solve the simultaneous inequalities:
Either $x^2+2(1-3a)x-(1+5a) > 0$ and $x+a+1<0$,
or $x^2+2(1-3a)x-(1+5a) < 0$ and $x+a+1>0$.
</s>