y^(4/3)-13y^(2/3)+36=0

Solution for y^(4/3)-13y^(2/3)+36=0 equation:

y in (-oo:+oo)

y^(4/3)-(13*y^(2/3))+36 = 0

y^(4/3)-13*y^(2/3)+36 = 0

t_1 = y^(2/3)

1*t_1^2-13*t_1^1+36 = 0

t_1^2-13*t_1+36 = 0

DELTA = (-13)^2-(1*4*36)

DELTA = 25

DELTA > 0

t_1 = (25^(1/2)+13)/(1*2) or t_1 = (13-25^(1/2))/(1*2)

t_1 = 9 or t_1 = 4

t_1 = 4

y^(2/3)-4 = 0

1*y^(2/3) = 4 // : 1

y^(2/3) = 4

y^(2/3) = 4 // ^ 3

y^2 = 64 // ^ 1/2

abs(y) = 8

y = 8 or y = -8

t_1 = 9

y^(2/3)-9 = 0

1*y^(2/3) = 9 // : 1

y^(2/3) = 9

y^(2/3) = 9 // ^ 3

y^2 = 729 // ^ 1/2

abs(y) = 27

y = 27 or y = -27

y in { 8, -8, 27, -27 }