all i can say is that its not going to be just solving it man...
its a pretty long process.
Its not difficult but it does take time. anyway this is what you need to do:
First you need to read up on depressed quartic equations. Once you have make your equation into a depressed quartic.
After youve done that you need to get familiar with Ferrari's solution. It has quite a few outcomes depending on the initial conditions so it might be worth taking a note of that when you show the different methods of solving the polynomial.
It turns out that solving a quartic boils down to solving a cubic.
We have . Let the roots of be .
Under the substitution the quartic becomes , where .
The roots of are . Let's call these respectively.
Now define . Direct verification shows the roots of are
So we have from above and from we get since the sum of the roots will give the coefficient of the cubic term, and in this case it's zero.
This gives us and .
Similarly
and
and .
Also an easy computation shows , so the choice of two of the square roots determines the third. Using some algebra we get
From above, knowing the 's lets us know the 's. Thus we've found the roots of .
And here's the final result after making the substitutions back to : quartic formula.