1/3y^3-1/2y^2=-1/6

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Solution for 1/3y^3-1/2y^2=-1/6 equation: 


y in (-oo:+oo)

(1/3)*y^3-((1/2)*y^2) = -1/6 // + -1/6

(1/3)*y^3-((1/2)*y^2)-(-1/6) = 0

(1/3)*y^3+(-1/2)*y^2+1/6 = 0

(1/3)*y^3+(-1/2)*y^2+1/6 = 0

(1/3)*y^3+(-1/2)*y^2+1/6 = 0 // * 0

{ 1, -1 }

1

y = 1

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

1

y-1

2*y^2-y-1

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

y-1

2*y^2-2*y^3

1-y^2

y^2-y

1-y

y-1

0

2*y^2-y-1 = 0

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

DELTA = 9

DELTA > 0

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

y = 1 or y = -1/2

y in { -1/2, 1, 1} 

y in { -1/2, 1, 1 }

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