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

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

y in (-oo:+oo)

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

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

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

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

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

4*y^2-15*y+9 = 0

4*y^2-15*y+9 = 0

4*y^2-15*y+9 = 0

DELTA = (-15)^2-(4*4*9)

DELTA = 81

DELTA > 0

y = (81^(1/2)+15)/(2*4) or y = (15-81^(1/2))/(2*4)

y = 3 or y = 3/4

(y-3/4)*(y-3) = 0

((y-3/4)*(y-3))/3 = 0

((y-3/4)*(y-3))/3 = 0 // * 3

(y-3/4)*(y-3) = 0

( y-3/4 )

y-3/4 = 0 // + 3/4

y = 3/4

( y-3 )

y-3 = 0 // + 3

y = 3

y in { 3/4, 3 }