(10n^2-35n-20)/(n-4)

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Solution for (10n^2-35n-20)/(n-4) equation: 


D( n )

n-4 = 0

n-4 = 0

n-4 = 0

n-4 = 0 // + 4

n = 4

n in (-oo:4) U (4:+oo)

(10*n^2-(35*n)-20)/(n-4) = 0

(10*n^2-35*n-20)/(n-4) = 0

10*n^2-35*n-20 = 0

5*(2*n^2-7*n-4) = 0

2*n^2-7*n-4 = 0

DELTA = (-7)^2-(-4*2*4)

DELTA = 81

DELTA > 0

n = (81^(1/2)+7)/(2*2) or n = (7-81^(1/2))/(2*2)

n = 4 or n = -1/2

5*(n+1/2)*(n-4) = 0

5*(n+1/2) = 0

( 5 )

5 = 0

n belongs to the empty set

( n+1/2 )

n+1/2 = 0 // - 1/2

n = -1/2

n = -1/2

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