Notes for AKT-140314/0:02:29

Boaz wrote:

I have one comment- you claim it is weaker than Galois. At least in one
important aspect I think it is much stronger. I challenge you to show with
Galois theory that if we add exp(z) to our list of allowed operations, a
solution cannot be found...
In fact, I guess you can add to the list of allowed functions any continuous
multivalued function with up to 4 values and it won't help (of course with 5
values you can solve it since the solution is such a function...). Am I right?

I believe Boaz is right, and he is making a very strong point - not only there is no formula for the roots of a quadratic in terms of radicals - in fact this remains true even if we are allowed to use other univalent functions.