How to solve this summation/polynomial? Is there a simple/fast way to find $y$ for the equation:
$$120,000=8000\sum_{t=1}^{4}\frac{1}{(1+y)^t}+\frac{100,000}{(1+y)^4}$$
?
I am trying to calculate the yield to maturity of a bond, and the answer is 2.66% or 2.67% (depending on your rounding off). I know some other method (some sort of trail and run method), but its rather long in my opinion.
The question was:

A bond has an annual coupon (interest) rate of 8%,
  with nominal value of 100,000 that has maturity in 4 years time. If
  the bond sells at 120,000, what is the yield to maturity?

 A: Hint: use $1+a+a^2+..+a^k = \displaystyle\frac{a^{k+1}-1}{a-1}$ for $b:=\displaystyle\frac1{1+y}$
A: Remember that $$\sum_{k=0}^{n+1}x^k=\frac{1-x^n}{1-x}$$
So: 
$$\begin{align*}120 =& 8\sum_{t=1}^{4}\frac{1}{(1+y)^t}+\frac{100}{(1+y)^4}=8\left(\frac{1-\frac{1}{(1+y)^5}}{1-\frac{1}{1+y}}-1\right)+\frac{100}{(1+y)^4}\\
=&8\frac{(1+y)^5-1-(1+y)^5+(1+y)^4}{(1+y)^5-(1+y)^4}+\frac{100}{(1+y)^4}\\
=&\frac{8(1+y)^4-8+100(1+y-1)}{(1+y)^4(1+y-1)}=\frac{8(1+y)^4-8+100y}{(1+y)^4y}\end{align*}$$
Multiplying and both sides by $y(1+y)^4$, dividing by $4$ and rearranging, we get:
$$(30y-2)(1+y)^4-25y+2=0$$
A: Either let $z=1+y$ and multiply through by $z^4$, or let $z=\frac{1}{1+y}$ and do nothing. Your equation is a quartic equation in $z$.
There is a closed form formula for the roots of the quartic, obtained initially by Cardano and Ferrari in the sixteenth century.  Variants were found by several mathematicians, including Descartes, Newton, and Lagrange. All the formulas are  very complicated, and not suitable for your purposes.  The link above is meant only to show you how complicated the Cardano-Ferrari formula is. 
It is best to use a numerical method, such as the Newton-Raphson method. Special algorithms have also been developed for equations that arise from interest rate calculations. In your case, you will be able to supply a good initial estimate $z_0$ of $z$, so Newton-Raphson will converge very rapidly. A couple of iterations will suffice for an answer that is accurate enough for all practical purposes. 
