How do I solve this inequality with absolute value? $$|8x-x^2|>x-18$$
Steps I took:
$$8x-x^{ 2 }>x-18\quad \quad \quad 8x-x^{ 2 }<18-x\quad $$
$$-x^{ 2 }+7x+18>0\quad \quad \quad -x^{ 2 }+9x-18<0\quad $$
$$x^{ 2 }-7x-18<0\quad \quad \quad x^{ 2 }-9x+18>0\quad $$
$$(x-9)(x+2)<0\quad \quad \quad (x-6)(x-3)>0\quad $$
How do I go from here?
No actual solution, please. 
 A: Note that if $x < 18$ then the equation is automatically true, so you only need
consider $x \ge 18$.
In this case we have $x \ge 0$ and $8-x \le 0$, you can write the above as
$x (x-8) > x-18$, or
equivalently you need to figure out for what values of $x \ge 18$ is
$(x-3)(x-6) >0$?
A: Hint: You now have an expression for the explicit zeros. So, in between the two zeros on each side of the equation (as well as less than all the zeros and greater than all of the zeros), the function can only be positive or negative (since it doesn't cross the $x$-axis). Plug in test points and check whether it is positive or negative there. Can you follow through the rest?
A: Hint:
A general and standard way to solve this kind of equations proceeds with the steps:
1) find the intervals where the argument of the absolute value is positive or negative. In this case we have: 
$$
8x-x^2>0 \iff 0<x<8 \qquad \land \qquad 8x-x^2>0 \iff x<0 \lor x>8
$$
2) Now we can split the equation in two systems:
$$
\begin{cases}
0<x<8\\
8x-x^2>x-18
\end{cases} \quad
\lor \quad
\begin{cases}
x<0 \lor x>8\\
x^2-8x>x-18
\end{cases}
$$
3) solve the systems an take the union of the solution intervals.
Can you do this?
