You cannot find a min and max from the first equation. If you take the left hand side as a function, we can find a min and max. First you set the derivative equal to 0. This gives a quadratic equation that you can solve to find the relative min/maxes.
Inflection points are points where the 2nd derivative changes sign. Thus the 2nd derivative is 0 at that point. So you have to check both solutions of your quadratic second derivative. Looking at a parabola, if there really are two 0s of the second derivative then they both represent sign changes of the second derivative.
The first equation is actually the first derivative from a quartic equation. So with how that first derivative is set up there is no way to find min/max? Is that because there is no way to get x by itself?
And the second derivative I just do the quadratic function solutions (both + and -) to find the inflection points?
Oh I see what you are saying. You are correct you have to solve that cubic equation. What I would do is have something like wolfram-alpha solve it for you.
And the second one, you are on the right track. If your parabola has 2 solutions, each is an inflection point. If it has 1 or 0 solutions, neither is an inflection point. This is because the concavity (second derivative) never changes signs for those case.