(2z^3-6z^2)/(z^2-3z)

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Solution for (2z^3-6z^2)/(z^2-3z) equation: 


D( z )

z^2-(3*z) = 0

z^2-(3*z) = 0

z^2-(3*z) = 0

z^2-3*z = 0

z^2-3*z = 0

DELTA = (-3)^2-(0*1*4)

DELTA = 9

DELTA > 0

z = (9^(1/2)+3)/(1*2) or z = (3-9^(1/2))/(1*2)

z = 3 or z = 0

z in (-oo:0) U (0:3) U (3:+oo)

(2*z^3-(6*z^2))/(z^2-(3*z)) = 0

(2*z^3-6*z^2)/(z^2-3*z) = 0

2*z^3-6*z^2 = 0

2*z^2*(z-3) = 0

z-3 = 0 // + 3

z = 3

2*z^2*(z-3) = 0

z^2-3*z = 0

z*(z-3) = 0

z-3 = 0 // + 3

z = 3

z*(z-3) = 0

(2*z^2*(z-3))/(z*(z-3)) = 0

( 2*z^2 )

2*z^2 = 0 // : 2

z^2 = 0

z = 0

( z-3 )

z-3 = 0 // + 3

z = 3

z in { 0} 

z in { 3} 

z belongs to the empty set

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