7x^(2/3)-34x^(1/3)-5=0

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Solution for 7x^(2/3)-34x^(1/3)-5=0 equation: 


x in (-oo:+oo)

7*x^(2/3)-(34*x^(1/3))-5 = 0

7*x^(2/3)-34*x^(1/3)-5 = 0

t_1 = x^(1/3)

7*t_1^2-34*t_1^1-5 = 0

7*t_1^2-34*t_1-5 = 0

DELTA = (-34)^2-(-5*4*7)

DELTA = 1296

DELTA > 0

t_1 = (1296^(1/2)+34)/(2*7) or t_1 = (34-1296^(1/2))/(2*7)

t_1 = 5 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 = 5

x^(1/3)-5 = 0

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

x^(1/3) = 5

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

x = 125

x = 125

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