7x^(2/3)-41x^(1/3)-6=0

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Solution for 7x^(2/3)-41x^(1/3)-6=0 equation: 


x in (-oo:+oo)

7*x^(2/3)-(41*x^(1/3))-6 = 0

7*x^(2/3)-41*x^(1/3)-6 = 0

t_1 = x^(1/3)

7*t_1^2-41*t_1^1-6 = 0

7*t_1^2-41*t_1-6 = 0

DELTA = (-41)^2-(-6*4*7)

DELTA = 1849

DELTA > 0

t_1 = (1849^(1/2)+41)/(2*7) or t_1 = (41-1849^(1/2))/(2*7)

t_1 = 6 or t_1 = -1/7

t_1 = -1/7

x^(1/3)+1/7 = 0

1*x^(1/3) = -1/7 // : 1

x^(1/3) = -1/7

( -1/7 < 0 i 1/3 in (0:1) ) => x należy do O

t_1 = 6

x^(1/3)-6 = 0

1*x^(1/3) = 6 // : 1

x^(1/3) = 6

x^(1/3) = 6 // ^ 3

x = 216

x = 216

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