x-((x^3-2x^2-5)/(3x^2-4x))

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Solution for x-((x^3-2x^2-5)/(3x^2-4x)) equation:


D( x )

3*x^2-(4*x) = 0

3*x^2-(4*x) = 0

3*x^2-(4*x) = 0

3*x^2-4*x = 0

3*x^2-4*x = 0

DELTA = (-4)^2-(0*3*4)

DELTA = 16

DELTA > 0

x = (16^(1/2)+4)/(2*3) or x = (4-16^(1/2))/(2*3)

x = 4/3 or x = 0

x in (-oo:0) U (0:4/3) U (4/3:+oo)

x-((x^3-(2*x^2)-5)/(3*x^2-(4*x))) = 0

x-((x^3-2*x^2-5)/(3*x^2-4*x)) = 0

(-1*(x^3-2*x^2-5))/(3*x^2-4*x)+x = 0

3*x^2-4*x = 0

3*x^2-4*x = 0

x*(3*x-4) = 0

3*x-4 = 0 // + 4

3*x = 4 // : 3

x = 4/3

x*(x-4/3) = 0

(-1*(x^3-2*x^2-5))/(x*(x-4/3))+x = 0

(-1*(x^3-2*x^2-5))/(x*(x-4/3))+(x^2*(x-4/3))/(x*(x-4/3)) = 0

x^2*(x-4/3)-1*(x^3-2*x^2-5) = 0

2/3*x^2+5 = 0

(2/3*x^2+5)/(x*(x-4/3)) = 0

(2/3*x^2+5)/(x*(x-4/3)) = 0 // * x*(x-4/3)

2/3*x^2+5 = 0

2/3*x^2 = -5 // : 2/3

x^2 = -15/2

x belongs to the empty set

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