x^(2/5)+7x^(1/5)+12=0

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Solution for x^(2/5)+7x^(1/5)+12=0 equation: 


x in (-oo:+oo)

x^(2/5)+7*x^(1/5)+12 = 0

t_1 = x^(1/5)

1*t_1^2+7*t_1^1+12 = 0

t_1^2+7*t_1+12 = 0

DELTA = 7^2-(1*4*12)

DELTA = 1

DELTA > 0

t_1 = (1^(1/2)-7)/(1*2) or t_1 = (-1^(1/2)-7)/(1*2)

t_1 = -3 or t_1 = -4

t_1 = -4

x^(1/5)+4 = 0

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

x^(1/5) = -4

( -4 < 0 i 1/5 in (0:1) ) => x naleu017Cy do O

t_1 = -3

x^(1/5)+3 = 0

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

x^(1/5) = -3

( -3 < 0 i 1/5 in (0:1) ) => x naleu017Cy do O

x belongs to the empty set

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