# How do I solve for n: $\dfrac{2}{9}(n^2-n)\times0.6^n=0.121$

So I need to solve this equation as a part of a binomial distribution problem:

Complete Question: Over a one moth period, Ava and Sven play a total of x games of tennis. The probability that Ava wins any game $0.4$. The result of each game played is independent of any other game played. Let X denote the number of games won by Ava over a one month period.

(a) Find an expression for $P(X = 2)$ in terms of $n$.

(b) If the probability that Ava wins two games in 0.121 correct to 3 d.p., find the value on $n$.

$\dfrac{2}{9}(n^2-n)\times0.6^n=0.121$

I've tried to simplify it by multiplying both sides by $\dfrac{9}{2}$:

$$(n^2-n)\times0.6^n=0.5445$$ Now I'm stuck, I tried using logarithms but I didn't get anywhere. The answer should be very close to $10$. Does anyone have a hint on how to proceed? Thank you.

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W|A can't solve it... (algebraically) [here] (wolframalpha.com/input/?i=solve+%28x^2-x%29*.6^x+%3D+0.5445) –  anorton Feb 25 '13 at 21:14
@anorton So what do you think I should do? This question is a medium difficulty one, and I am sure the equation is right, because it was part a of the question. –  Hans Groeffen Feb 25 '13 at 21:15
You're absolutely sure you're supposed to solve the equation for $n$? –  anorton Feb 25 '13 at 21:17
are you sure its 0.6^n and not 0.6^2n? –  MathGeek Feb 25 '13 at 21:19
homework should not be used as a standalone tag; see tag-wiki and meta. –  Martin Sleziak Feb 26 '13 at 7:32