>>9
What did I do that was sneaky? I expanded the parentheses (which I should have done with Pascal's Triangle instead of the Distributive Property), subtracted 1 from both sides, and factored out x from the expression containing it. That left me with x*[quadratic], which made 0 one root by default. Then, I used the Quadratic Formula to find the two roots of the remaining quadratic, and came up with -1.5 +/- i*sqrt(3/4) for the other two roots. There's nothing hard about the problem.
Algebra II, I think? In any case, it was a very straightforward and easy problem.