(y^2-4/9y+27)(y+3/y-2)=0

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Solution for (y^2-4/9y+27)(y+3/y-2)=0 equation: 


D( y )

y = 0

y = 0

y = 0

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

(y^2-((4/9)*y)+27)*(y+3/y-2) = 0

(y^2+(-4/9)*y+27)*(y+3/y-2) = 0

(y^2-4/9*y+27)*(y+3*y^-1-2) = 0

y^2-4/9*y+27 = 0

y^2-4/9*y+27 = 0

DELTA = (-4/9)^2-(1*4*27)

DELTA = -8732/81

DELTA < 0

1 = 0

y+3*y^-1-2 = 0

1*y^1+3*y^-1-2*y^0 = 0

(1*y^2-2*y^1+3*y^0)/(y^1) = 0 // * y^2

y^1*(1*y^2-2*y^1+3*y^0) = 0

y^1

y^2-2*y+3 = 0

y^2-2*y+3 = 0

DELTA = (-2)^2-(1*3*4)

DELTA = -8

DELTA < 0

y in { } 

1 = 0

1 = 0

y belongs to the empty set

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