5=-12/x-1/x^2

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Solution for 5=-12/x-1/x^2 equation: 


D( x )

x = 0

x^2 = 0

x = 0

x = 0

x^2 = 0

x^2 = 0

1*x^2 = 0 // : 1

x^2 = 0

x = 0

x in (-oo:0) U (0:+oo)

5 = -12/x-(1/(x^2)) // + -12/x-(1/(x^2))

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

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

12*x^-1+x^-2+5 = 0

t_1 = x^-1

1*t_1^2+12*t_1^1+5 = 0

t_1^2+12*t_1+5 = 0

DELTA = 12^2-(1*4*5)

DELTA = 124

DELTA > 0

t_1 = (124^(1/2)-12)/(1*2) or t_1 = (-124^(1/2)-12)/(1*2)

t_1 = (2*31^(1/2)-12)/2 or t_1 = (-2*31^(1/2)-12)/2

t_1 = (-2*31^(1/2)-12)/2

x^-1-((-2*31^(1/2)-12)/2) = 0

1*x^-1 = (-2*31^(1/2)-12)/2 // : 1

x^-1 = (-2*31^(1/2)-12)/2

-1 < 0

1/(x^1) = (-2*31^(1/2)-12)/2 // * x^1

1 = ((-2*31^(1/2)-12)/2)*x^1 // : (-2*31^(1/2)-12)/2

2*(-2*31^(1/2)-12)^-1 = x^1

x = 2*(-2*31^(1/2)-12)^-1

t_1 = (2*31^(1/2)-12)/2

x^-1-((2*31^(1/2)-12)/2) = 0

1*x^-1 = (2*31^(1/2)-12)/2 // : 1

x^-1 = (2*31^(1/2)-12)/2

-1 < 0

1/(x^1) = (2*31^(1/2)-12)/2 // * x^1

1 = ((2*31^(1/2)-12)/2)*x^1 // : (2*31^(1/2)-12)/2

2*(2*31^(1/2)-12)^-1 = x^1

x = 2*(2*31^(1/2)-12)^-1

x in { 2*(-2*31^(1/2)-12)^-1, 2*(2*31^(1/2)-12)^-1 }

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