I am new to economics maths and I have to solve some issues.
I am trying to solve the following equation : $$-156000 + \frac{57080}{(1+irr)^1} + \frac{81080}{(1+irr)^2} + \frac{176480}{(1+irr)^3} + \frac{213680}{(1+irr)^4} + \frac{190280}{(1+irr)^5} = 0$$
I solved the equation in Excel and the result is $0.8976$.
Could you help me please do it manually?
Thanks in advance.
Let $x=irr$. From the formula you have, we let $$ f(x)=-156000 + \frac{57080}{(1+x)^1} + \frac{81080}{(1+x)^2} + \frac{176480}{(1+x)^3} + \frac{213680}{(1+x)^4} + \frac{190280}{(1+x)^5}. $$ So, we see that this comes from initial invest $156000$ and getting back $57080$, $\ldots$, etc on the end of $1st$ through $5th$ years respectively. If the cash flow requires only one time investment and getting enough cash back such as this case, the function $f(x)$ satisfies the following:
$f(x)$ is continuous and decreasing for $x\geq 0$.
$f(0)>0$ and $\lim\limits_{x\rightarrow\infty} f(x) <0$.
$f(x)=0$ has a unique positive root.
The reason for 3. is by Intermediate Value Theorem.
As for obtaining a numerical value of this unique root, again refer to the Intermediate Value Theorem (IVT). If as you said, $0.6189$ is a root, then we will be able to see that $$ f(0.6)>0, \ f(0.7)<0. $$ Then by the IVT, we have $0.6<x<0.7$. If you want to narrow down this, the next one you try is $f(0.65)$. If this gives $f(0.65)<0$, then $0.6<x<0.65$. The next one you try is $f(0.625)$, etc.
So, this way of solving for the Internal Rate of Return takes too much time, but it is a good way if the calculator has only basic features such as plugging in numbers and doing arithmetics.
As other comments suggested, Wolfram Alpha or Excel is certainly a better choice if it is available. If you are able to use a Financial Calculator such as BA-IIplus, then there is an option for you to solve for IRR by plugging in the cashflow.