How do you solve equations of any degree? I have stuck solving this problem of financial mathematics, in this equation:
$$\frac{(1+x)^{8}-1}{x}=11$$
I'm stuck in this eight grade equation:
$$x^{8}+8x^{7}+28x^{6}+56x^{5}+70x^{4}+56x^{3}+28x^{2}+9x-11=0$$
But I cannot continue past this. This quickly lead me to find a general way of solving equations of any degree, but I couldn't find anything serious on the internet.
Do you know any simple methods to solve equations of any degree?
 A: As has been pointed out above, from Galois theory in general there is no algebraic solution to polynomials of degree 5+. 
But what about numerical solvers? Consider the polynomial 
$$p(x)=a+bx+cx^2+dx^3+x^4$$
and its companion matrix,
$$A(p)=\left[\begin{matrix}0 & 0 & 0 & -a \\ 1 & 0 & 0 & -b \\ 0 & 1 & 0 & -c \\ 0 & 0 & 1 & -d \end{matrix}\right]$$
If you stare at it long enough, hopefully you can convince yourself that the eigenvalues of the companion matrix are exactly the zeros of the original polynomial. (hint, what is $\det(A-xI)$?)
So finding the zeros of a polynomial is equivalent to solving an eigenvalue problem, which is well studied numerically.
If you want to find a single one of the roots, consider using some form of inverse iteration/Rayleigh quotient iteration. If you want to find all the roots, consider using some variant of the QR algorithm (not to be confused with QR decomposition, which is used inside the QR algorithm).
A: First off, let me say that the polynomial you gave does not match up with your original problem.  So, I will solve it based on your original problem and not the polynomial you gave.  Perhaps you typed it wrong.
Second off, since this is financial mathematics, there are most likely assumptions that go beyond the pure math.  The point is, you don't need to know general methods for solving any polynomial equation.  You need methods for very specific types of situations.
$x$ here represents an interest rate.  And, it is often assumed that interest rates are positive.  This simplifies the solution considerably.  But, even without that assumption, I can tell by looking at your equation what your original problem was.  Your original problem was that 1 dollar is put into an account at the end of each year for 8 years.  The account accumulates at interest rate $x$, or $100x\%$, and the accumulated value at time 8 is 11.  Since you end up with more money than you put in, it is definitely reasonable to assume your interest rate is positive.  In fact, since all the payments by you were done ahead of the payment at time 8 by the bank of 11, there is guaranteed to be only one real solution, and it is positive clearly from the context.
So, now your job is just to figure out the one and only solution.  We use the Intermediate Value Theorem.  It says, if a function is continuous on some interval, and if the function is negative at one point and positive at another point, then it must be 0 somewhere in between.  So consider
$$f(x) = \frac{(1 + x)^8 - 1}{x} - 11.$$
It is not continuous everywhere, because of division by 0 when $x = 0$, but it is continuous for $x > 0$ and therefore the Intermediate Value Theorem applies if we are only searching for a solution to $f(x) = 0$ for positive $x$.
Plug in a couple small numbers, like 0.05 and 0.2.  $f(0.05) = -1.45089$ and $f(0.2) = 5.49908$.  Now, we know the solution is between 0.05 and 0.2 by the Intermediate Value Theorem.  Pick a number in between, like 0.1.  $f(0.1) = 0.435888$.  Since this is positive, our root is between 0.05 and 0.1.  We're getting close.  Try 0.075. $f(0.075) < 0$.  This tells us our solution is between 7.5% and 10%. Try 0.09.  $f(0.09) > 0$, so our solution is between 0.075 and 0.09.  Try 0.08.  You get $f(0.08) = -0.363372$.  Continue in this manner and you'll eventually narrow down the answer to as many decimal places as you want.  Of course, it's slow.  This method is basically the bisection method, except I didn't always pick my next number to be right in the middle of the previous bounds.  You could also try Newton's method.  You'll eventually get x = 0.0892863...
A: I don't know of a universal method to solve any equation. If you have the luxury to choose any method to solve an equation, and you are not particularly interested in advanced math, you could try to plot the equation either by hand or use free software on the net. to give you at least a starting solution.
For example, your equation 
$\frac{(1+x)^{8}-1}{x}=11$
looks like this when plotted. 
Using plots of this sort only shows real solution (not complex ones). In this case $x=.089311$ is a solution to the equation (good for 3 digits after the decimal point).
