(z-3)2/3=(7z-33)1/3

Solution for (z-3)2/3=(7z-33)1/3 equation:

z in (-oo:+oo)

((z-3)^2)/3 = ((7*z-33)^1)/3 // - ((7*z-33)^1)/3

((z-3)^2)/3-(((7*z-33)^1)/3) = 0

((z-3)^2)/3-((7*z-33)/3) = 0

((z-3)^2)/3+(-1*(7*z-33))/3 = 0

(z-3)^2-1*(7*z-33) = 0

z^2-13*z+42 = 0

z^2-13*z+42 = 0

z^2-13*z+42 = 0

DELTA = (-13)^2-(1*4*42)

DELTA = 1

DELTA > 0

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

z = 7 or z = 6

(z-6)*(z-7) = 0

((z-6)*(z-7))/3 = 0

((z-6)*(z-7))/3 = 0 // * 3

(z-6)*(z-7) = 0

( z-6 )

z-6 = 0 // + 6

z = 6

( z-7 )

z-7 = 0 // + 7

z = 7

z in { 6, 7 }