(x-3)^2/5=4x^1/5

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


x in (-oo:+oo)

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

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

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

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

(x-3)^2-4*x = 0

x^2-10*x+9 = 0

x^2-10*x+9 = 0

x^2-10*x+9 = 0

DELTA = (-10)^2-(1*4*9)

DELTA = 64

DELTA > 0

x = (64^(1/2)+10)/(1*2) or x = (10-64^(1/2))/(1*2)

x = 9 or x = 1

(x-1)*(x-9) = 0

((x-1)*(x-9))/5 = 0

((x-1)*(x-9))/5 = 0 // * 5

(x-1)*(x-9) = 0

( x-1 )

x-1 = 0 // + 1

x = 1

( x-9 )

x-9 = 0 // + 9

x = 9

x in { 1, 9 }

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