Prove that $\displaystyle f(x) = 4x^4+2x^2-6x+2=0$ has exactly two solutions.

Proof so far.

Now $\displaystyle f'(x)=4x^3+4x-6$, set it equals to zero, we have $\displaystyle 2x^3+2x-3=0$, so this equation have 3 solutions, meaning we have 3 extremas, meaning the equation can cross the x-axis twice.

Would that be enough to prove it?