4x^(2/3)-25x^(1/3)+25=0

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Solution for 4x^(2/3)-25x^(1/3)+25=0 equation: 


x in (-oo:+oo)

4*x^(2/3)-(25*x^(1/3))+25 = 0

4*x^(2/3)-25*x^(1/3)+25 = 0

t_1 = x^(1/3)

4*t_1^2-25*t_1^1+25 = 0

4*t_1^2-25*t_1+25 = 0

DELTA = (-25)^2-(4*4*25)

DELTA = 225

DELTA > 0

t_1 = (225^(1/2)+25)/(2*4) or t_1 = (25-225^(1/2))/(2*4)

t_1 = 5 or t_1 = 5/4

t_1 = 5/4

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

1*x^(1/3) = 5/4 // : 1

x^(1/3) = 5/4

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

x = 125/64

t_1 = 5

x^(1/3)-5 = 0

1*x^(1/3) = 5 // : 1

x^(1/3) = 5

x^(1/3) = 5 // ^ 3

x = 125

x in { 125/64, 125 }

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