y^3-28y^(3/2)+27=0

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Solution for y^3-28y^(3/2)+27=0 equation: 


D( y )

y < 0

y < 0

y in <0:+oo)

y^3-(28*y^(3/2))+27 = 0

y^3-28*y^(3/2)+27 = 0

t_1 = y^(3/2)

1*t_1^2-28*t_1^1+27 = 0

t_1^2-28*t_1+27 = 0

DELTA = (-28)^2-(1*4*27)

DELTA = 676

DELTA > 0

t_1 = (676^(1/2)+28)/(1*2) or t_1 = (28-676^(1/2))/(1*2)

t_1 = 27 or t_1 = 1

t_1 = 1

y^(3/2)-1 = 0

1*y^(3/2) = 1 // : 1

y^(3/2) = 1

y^(3/2) = 1 // ^ 2

y^3 = 1 // ^ 1/3

y = 1

t_1 = 27

y^(3/2)-27 = 0

1*y^(3/2) = 27 // : 1

y^(3/2) = 27

y^(3/2) = 27 // ^ 2

y^3 = 729 // ^ 1/3

y = 9

y in { 1, 9 }

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