Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$
The left-hand side of that [equation](https://artofproblemsolving.com/wiki/index.php/Equation) is nearly equal to $(x - 1)^4$.  Thus, we add 1 to each side in order to complete the fourth power and get
$(x - 1)^4 = 2006$.
Let $r = \sqrt[4]{2006}$ be the positive [real](https://artofproblemsolving.com/wiki/index.php/Real) fourth root of 2006.  Then the roots of the above equation are $x = 1 + i^n r$ for $n = 0, 1, 2, 3$.  The two non-real members of this set are $1 + ir$ and $1 - ir$.  Their product is $P = 1 + r^2 = 1 + \sqrt{2006}$.  $44^2 = 1936 < 2006 < 2025 = 45^2$ so $\lfloor P \rfloor = 1 + 44 = \boxed{045}$.