solve $|x-6|>|x^2-5x+9|$ 
solve $|x-6|>|x^2-5x+9|,\ \ x\in \mathbb{R}$

I have done $4$ cases.
$1.)\ x-6>x^2-5x+9\ \ ,\implies  x\in \emptyset \\
2.)\ x-6<x^2-5x+9\ \ ,\implies x\in \mathbb{R} \\
3.)\  -(x-6)>x^2-5x+9\ ,\implies 1<x<3\\
4.)\  (x-6)>-(x^2-5x+9),\ \implies x>3\cup x<1 $
I am confused on how I proceed.
Or if their is any other short way than making $4$ cases than I would like to know. 
I have studied maths up to $12$th grade. Thanks.

HINT:
As $x^2-5x+9=\dfrac{(2x-5)^2+11}4\ge\dfrac{11}4>0\forall $ real $x$ 
So, $|x^2-5x+9|=+(x^2-5x+9)\forall $ real $x$ 
So, the problem reduces to two cases for $|x-6|$
Also, $|x-6|>\dfrac{11}4\implies \pm(x-6)>\dfrac{11}4$
