Periodic and odd function

Question

Problem from Georgian National Exam (Students know only pre-algebra)

Odd function $f$ with period $4$ is determined on the set of the real numbers. On the interval [0;2] $f$ function is given by the equation: $f(x) = 4x-2x^2 $. Find the set of solutions of the equation: $2f(x)\cdot f(x-8) - 3f(x+12) - 2 = 0$

Answer 1

So you substituted $$f(x-8)=f(x+12)=f(x)=y$$ so you get $$2y^2-3y-2=0$$ Can you solve this? then once you have $y$ try to find the solution of $$4x-2x^2=y$$ So you have two $y$ values, which will have two $x$ values each. You choose the solution in the $[0,2]$ interval.