To solve the inequality log1/6(x^2-3x+2) >= -1, we first need to rewrite it in exponential form:

1/6^(log1/6(x^2-3x+2)) >= 1

This can be simplified further to:

(x^2-3x+2)^1/6 >= 1

Now, raise both sides to the power of 6:

(x^2-3x+2) >= 1

Now, solve the quadratic inequality:

x^2 - 3x + 2 - 1 >= 0
x^2 - 3x + 1 >= 0

Now, we can factor the quadratic equation:

(x-1)(x-1) >= 0

This simplifies to:

(x-1)^2 >= 0

Since the square of a number is always non-negative, the inequality holds true for all real numbers of x. Thus, the solution to the inequality log1/6(x^2-3x+2) >= -1 is all real numbers.