Find $2f(x)\cdot f(x-8) - 3f(x+12) - 2 = 0$ 
Function $f$ $\in \mathbb{R}$  is odd and has a period of $4$. On a $[0,2]$ segment function $f$ is defined as $f(x)= 4x - 2x^2$. Find the set of solutions for the equation: $$2f(x)\cdot f(x-8) - 3f(x+12) - 2 = 0$$

So, here's my attempt: function has a period of $4$ means that $f(x) = f(x+4) = f(x+8) =  f(x+12)$. Also, from the fact that $f$ is odd,  $f(x-8)$ just means $-f(x+8) = -f(x)$ (I might be wrong here).
Now let $4x-2x^2 = z$. We get the equation: $$2z\cdot(-z)-3z-2=0$$ $$-2z^2-3z-2=0$$ $$2z^2+3z+2 = 0 $$
$D < 0$, so this doesn't have a solution. What am I doing wrong?
 A: In your solution, you make the following error
$$f(x) \space \text{is odd} \implies f(x-8) = -f(x+8)$$
This is not true, $f$ being odd means that $f(-x) = -f(-x)$, hence $f(x-8) = -f(-x+8)$
Now, since the period is 4, what we can say is that $f(x-8) = f(x-4) = f(x)$
Hence, replacing $f(x) = z$
$$2z^2 - 3z-2 = 0$$
$$\implies z = \frac{3 \pm 5}{4} = -\frac{1}{2},2$$
Now is where your odd function restriction will come into play
EDIT
Adding as it is not straightforward how to continue. Since $f$ is only defined on $[0,2]$, you extend it to $[-2,2]$ using the odd restriction. Hence $f(x)$ is defined as
$$f(x) = \begin{cases}4x-2x^2 & x\in[0,2] \\ 4x+2x^2 & x \in[-2,0]\end{cases}$$
Now, if you solve on this full domain for the above, you have
$$4x+2x^2 = -\frac{1}{2} \implies x = -1 \pm \frac{\sqrt{3}}{2}$$
$$4x - 2x^2 = 2 \implies x = 1$$
Now, since our full function $f$ is a periodic repetition of this, the roots will get repeated as well, and get shifted either up or down by 4 units each. Hence the roots will be of the form
$$x = \begin{cases}1 + 4k \\ 4k-1 + \frac{\sqrt{3}}{2} \\ 4k - 1 - \frac{\sqrt{3}}{2}\end{cases}$$
