4x^(2/3)-19x^(1/3)-5=0

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Solution for 4x^(2/3)-19x^(1/3)-5=0 equation: 


x in (-oo:+oo)

4*x^(2/3)-(19*x^(1/3))-5 = 0

4*x^(2/3)-19*x^(1/3)-5 = 0

t_1 = x^(1/3)

4*t_1^2-19*t_1^1-5 = 0

4*t_1^2-19*t_1-5 = 0

DELTA = (-19)^2-(-5*4*4)

DELTA = 441

DELTA > 0

t_1 = (441^(1/2)+19)/(2*4) or t_1 = (19-441^(1/2))/(2*4)

t_1 = 5 or t_1 = -1/4

t_1 = -1/4

x^(1/3)+1/4 = 0

1*x^(1/3) = -1/4 // : 1

x^(1/3) = -1/4

( -1/4 < 0 i 1/3 in (0:1) ) => x naleu017Cy 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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