|x-6|>|x2-5x+9|
x-6>0 when x>6.
x2-5x+9 has only complex zeroes and is always greater than zero. So the abs can be removed.
|x-6|>x2-5x+9.
When x>6: x-6>x2-5x+9 has to be solved. x2-6x+15<0 is false because this quadratic has only complex zeroes and is always greater than zero. Therefore we need to consider x-6<0.
6-x>x2-5x+9, x2-4x+3<0, (x-1)(x-3)<0 only when 1<x<3, which is the solution.