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

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


z in (-oo:+oo)

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

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

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

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

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

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

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

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

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

DELTA = 81

DELTA > 0

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

z = 3 or z = 3/4

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

((z-3/4)*(z-3))/5 = 0

((z-3/4)*(z-3))/5 = 0 // * 5

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

( z-3/4 )

z-3/4 = 0 // + 3/4

z = 3/4

( z-3 )

z-3 = 0 // + 3

z = 3

z in { 3/4, 3 }

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