6t^(2/5)+7t^(1/5)+2=0

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Solution for 6t^(2/5)+7t^(1/5)+2=0 equation: 


t in (-oo:+oo)

6*t^(2/5)+7*t^(1/5)+2 = 0

t_1 = t^(1/5)

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

6*t_1^2+7*t_1+2 = 0

DELTA = 7^2-(2*4*6)

DELTA = 1

DELTA > 0

t_1 = (1^(1/2)-7)/(2*6) or t_1 = (-1^(1/2)-7)/(2*6)

t_1 = -1/2 or t_1 = -2/3

t_1 = -2/3

t^(1/5)+2/3 = 0

1*t^(1/5) = -2/3 // : 1

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

( -2/3 < 0 i 1/5 in (0:1) ) => t naleu017Cy do O

t_1 = -1/2

t^(1/5)+1/2 = 0

1*t^(1/5) = -1/2 // : 1

t^(1/5) = -1/2

( -1/2 < 0 i 1/5 in (0:1) ) => t naleu017Cy do O

t belongs to the empty set

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