(5x-20)/(x^2-4x)

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Solution for (5x-20)/(x^2-4x) equation: 


D( x )

x^2-(4*x) = 0

x^2-(4*x) = 0

x^2-(4*x) = 0

x^2-4*x = 0

x^2-4*x = 0

DELTA = (-4)^2-(0*1*4)

DELTA = 16

DELTA > 0

x = (16^(1/2)+4)/(1*2) or x = (4-16^(1/2))/(1*2)

x = 4 or x = 0

x in (-oo:0) U (0:4) U (4:+oo)

(5*x-20)/(x^2-(4*x)) = 0

(5*x-20)/(x^2-4*x) = 0

x^2-4*x = 0

x*(x-4) = 0

x-4 = 0 // + 4

x = 4

x*(x-4) = 0

(5*x-20)/(x*(x-4)) = 0

5*x-20 = 0 // + 20

5*x = 20 // : 5

x = 20/5

x = 4

x in { 4} 

x belongs to the empty set

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