I was told in another post that Galois theory could help you solve solvable polynomials, and that solvable polynomials had roots that could be expressed as functions of rational numbers, but that it couldn't help me solve a polynomial with rational roots. It cannot help me solve polynomials with roots that couldn't be expressed with radicals. So doesn't that leave me with only polynomials with roots that ARE expressed in radical that Galois theory can help me solve? In other words, does a polynomial's roots HAVE to contain at least one radical for Galois theory to help you solve it?
Let's look at an example. Does Galois Theory help you solve $x^2-3x+1=0$? In a sense, it does; you could use Galois Theory to analyze that equation and eventually to write down its solutions. But you'd be insane to do that: it's ever so much simpler to just write down the answers given by the quadratic formula. Moreover, there is no way that someone who doesn't already know the quadratic formula is going to be able to understand the Galois Theory approach to this equation.
Now for a polynomial whose roots are all rational, it's even worse. The Galois group is trivial, and it basically tells you that the way to find the roots is to use the rational root theorem.
So I could say, yes, Galois Theory helps you solve the equation; it helps you by telling you to use the rational root theorem. You might reply, Galois Theory doesn't help you solve the equation; all it does it tell you to use the rational root theorem. We would agree on the mathematics, and disagree on the interpretation. I don't think we can take it much further than that.