Hi !
It's not easy for general formulae...
If you know one of the roots, then you will easily be able to find the others (polynomial factorisation).
Now, I know there is a general formula for
------------------
To refer to your exercise, you're looking for :
you can change x+a for y :
------------------
If , then there is a real solution (& 2 complex ones) :
Unfortunately, I can't remember who discovered it (he was an Italian guy, in the XVI th century if my memory is correct)... I had to give back the book to the library.
Edit : his name is Bombelli.
---------------------
I've just seen you can use the Cardan method or trigonometric method... Google for it, I don't have enough knowledge to talk about it.
See Cardano's method.
-Dan
If you look at the general form as you go along you will note that the derivation adds one extra variable to the solution, giving us the freedom to choose a single condition that is convenient for us. This is the condition that we choose. Once this substitution is made the equation falls into something like the form
This is a "tri-quadratic" equation, I guess you'd call it. We can easily solve this for p by using the quadratic formula to solve for then simply taking the cube root to find p. Then you can back substitute to get your answer in terms of the original coefficients.
You can get a tutorial on LaTeX here, in the first few posts of the sticky thread at the top of the list.
-Dan