4x+8/x-3=8

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Solution for 4x+8/x-3=8 equation: 


D( x )

x = 0

x = 0

x = 0

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

4*x+8/x-3 = 8 // - 8

4*x+8/x-8-3 = 0

4*x^1+8*x^-1-11*x^0 = 0

(4*x^2-11*x^1+8*x^0)/(x^1) = 0 // * x^2

x^1*(4*x^2-11*x^1+8*x^0) = 0

x^1

4*x^2-11*x+8 = 0

4*x^2-11*x+8 = 0

DELTA = (-11)^2-(4*4*8)

DELTA = -7

DELTA < 0

x in { } 

x belongs to the empty set

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