Find interest rate in a leasing problem 
A firm has entered into a 3-year lease agreement for an audio broadcasting system valued at $\$18'000$. The contract calls for 8 deferred payments each four months period of $\$2'803.33$ each and a redemption value of $\$1'000$. Determine at what four months period interest rate the transaction was made.


Solving the problem, I found the following equation
$$2'803.33 \cdot \frac{(1+i)^8 - 1}{i} \cdot \frac{1}{(1+i)^8} +  \frac{1'000}{(1+i)^9} = 18'000$$
where $i$ is the desired interest rate. I really have some trouble to find manually that the only real solution, assuming $i \neq 0$, is $i = 0.06$.
Any suggestions? Thanks in advance!
 A: You can substitute $1+i$ by $q$. The equation becomes
$$2'803.33 \cdot \frac{q^8 - 1}{q-1} \cdot \frac{1}{q^8} +  \frac{1'000}{q^9} = 18'000$$
Multiplying the equation by $q^9$
$$2'803.33 \cdot \frac{q^9 - q}{q-1} +  1000= 18'000\cdot q^9$$
Multiplying the equation by $q-1$
$$2'803.33 \cdot (q^9 - q)+  1000\cdot (q-1)= 18'000\cdot q^9\cdot (q-1)$$
Multiplying out the brackets
$$2'803.33 \cdot q^9 - 2'803.33 \cdot q+  1000\cdot q -1000= 18'000\cdot q^{10}-18'000\cdot q^9$$
Putting all terms on the right hand side
$$0= 18'000\cdot q^{10}-18'000\cdot q^9-2'803.33 \cdot q^9 +2'803.33 \cdot q-1000\cdot q+1000$$
$$0= 18'000\cdot q^{10}-20'803.33\cdot q^9 +1'803.33 \cdot q+1000$$
So basically you have to find the root(s) of a 10 degree polynomial. This equation can not solved algebraically.  You have to apply an approximation method, like the  Newton-Raphson method, to solve the equation. Intuitively I would have used the initial value of $q_0=1+i=1.05$
A: The equation of value is correct:  $$\require{enclose} 18000 = 2803.33 a_{\enclose{actuarial}{8} i} + 1000 v^9 = 2803.33 \frac{1-v^8}{i} + 1000 v^9,$$ where $v = 1/(1+i)$ is the $4$-month present value discount factor, and $i$ is the $4$-month effective rate of interest.  We can also write this as
$$18000 = 2803.33 \frac{v - v^9}{1-v} + 1000 v^9.$$  This gives the degree-$10$ polynomial
$$f(v) = v^{10} + 1.80333 v^9 - 20.80333 v + 18 = 0.$$  To find nontrivial real roots, we employ Newton's method:
$$v_{n+1} = v_n - \frac{f(v_n)}{f'(v_n)}$$ with the initial guess $v_0 = 0.9$, since $v = 1$ is a trivial root.  This gives the table
$$\begin{array}{c|c|c}
n & v_n & i_n \\
\hline
0 & 0.9 & 0.111111 \\
1 & 0.93262 & 0.0722484 \\
2 & 0.942127 & 0.0614279 \\
3 & 0.943373 & 0.0600266 \\
4 & 0.943396 & 0.0600003 \\
5 & 0.943396 & 0.0600003
\end{array}$$
