Question 268925
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Not sure what you want to solve.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(n)\ =\ -2n^2\ +\ 40n\ +\ 20]


Is a lovely little quadratic function.  But since you haven't provided a value for *[tex \LARGE P(n)], there are an infinite number of ordered pairs *[tex \LARGE \left(n,P(n)\right)] that satisfy the equation.


Now if what you really meant to ask is "What are the roots of *[tex \LARGE P(n)]?", in other words, what are the solutions to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2n^2\ +\ 40n\ +\ 20\ =\ 0] 


That is another thing altogether.  Presuming that is your real question, proceed as follows.


Multiply both sides by *[tex \Large -\frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n^2\ -\ 20n\ -\ 10\ =\ 0]


The quadratic does not factor, so use the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n = \frac{-(-20)\ \pm\ \sqrt{(-20)^2\ -\ 4(1)(-10)}}{2(1)} ]


You can do your own arithmetic.  Note that the prime factorization of 440 is 2 x 2 x 2 x 5 x 11, so the best you will be able to do with the radical is *[tex \LARGE 2\sqrt{110}]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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