y^(2/5)-10y^(1/5)+9=0

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Solution for y^(2/5)-10y^(1/5)+9=0 equation: 


y in (-oo:+oo)

y^(2/5)-(10*y^(1/5))+9 = 0

y^(2/5)-10*y^(1/5)+9 = 0

t_1 = y^(1/5)

1*t_1^2-10*t_1^1+9 = 0

t_1^2-10*t_1+9 = 0

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

DELTA = 64

DELTA > 0

t_1 = (64^(1/2)+10)/(1*2) or t_1 = (10-64^(1/2))/(1*2)

t_1 = 9 or t_1 = 1

t_1 = 1

y^(1/5)-1 = 0

1*y^(1/5) = 1 // : 1

y^(1/5) = 1

y^(1/5) = 1 // ^ 5

y = 1

t_1 = 9

y^(1/5)-9 = 0

1*y^(1/5) = 9 // : 1

y^(1/5) = 9

y^(1/5) = 9 // ^ 5

y = 59049

y in { 1, 59049 }

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