(3z-4)^2/5=(16z)^1/5

Solution for (3z-4)^2/5=(16z)^1/5 equation:

z in (-oo:+oo)

((3*z-4)^2)/5 = ((16*z)^1)/5 // - ((16*z)^1)/5

((3*z-4)^2)/5-(((16*z)^1)/5) = 0

((3*z-4)^2)/5+(-16/5)*z = 0

((3*z-4)^2)/5+(-16*z)/5 = 0

(3*z-4)^2-16*z = 0

9*z^2-40*z+16 = 0

9*z^2-40*z+16 = 0

9*z^2-40*z+16 = 0

DELTA = (-40)^2-(4*9*16)

DELTA = 1024

DELTA > 0

z = (1024^(1/2)+40)/(2*9) or z = (40-1024^(1/2))/(2*9)

z = 4 or z = 4/9

(z-4/9)*(z-4) = 0

((z-4/9)*(z-4))/5 = 0

((z-4/9)*(z-4))/5 = 0 // * 5

(z-4/9)*(z-4) = 0

( z-4/9 )

z-4/9 = 0 // + 4/9

z = 4/9

( z-4 )

z-4 = 0 // + 4

z = 4

z in { 4/9, 4 }