2n-9=35/n

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Solution for 2n-9=35/n equation: 


D( n )

n = 0

n = 0

n = 0

n in (-oo:0) U (0:+oo)

2*n-9 = 35/n // - 35/n

2*n-(35/n)-9 = 0

2*n-35*n^-1-9 = 0

2*n^1-35*n^-1-9*n^0 = 0

(2*n^2-9*n^1-35*n^0)/(n^1) = 0 // * n^2

n^1*(2*n^2-9*n^1-35*n^0) = 0

n^1

2*n^2-9*n-35 = 0

2*n^2-9*n-35 = 0

DELTA = (-9)^2-(-35*2*4)

DELTA = 361

DELTA > 0

n = (361^(1/2)+9)/(2*2) or n = (9-361^(1/2))/(2*2)

n = 7 or n = -5/2

n in { -5/2, 7} 

n in { -5/2, 7 }

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