(1/n+3)+(5/n^2-9)=(2/n-3)

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Solution for (1/n+3)+(5/n^2-9)=(2/n-3) equation: 


D( n )

n = 0

n^2 = 0

n = 0

n = 0

n^2 = 0

n^2 = 0

1*n^2 = 0 // : 1

n^2 = 0

n = 0

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

1/n+5/(n^2)-9+3 = 2/n-3 // - 2/n-3

1/n-(2/n)+5/(n^2)-9+3+3 = 0

1/n-2*n^-1+5/(n^2)-9+3+3 = 0

5*n^-2-n^-1-3 = 0

t_1 = n^-1

5*t_1^2-1*t_1^1-3 = 0

5*t_1^2-t_1-3 = 0

DELTA = (-1)^2-(-3*4*5)

DELTA = 61

DELTA > 0

t_1 = (61^(1/2)+1)/(2*5) or t_1 = (1-61^(1/2))/(2*5)

t_1 = (61^(1/2)+1)/10 or t_1 = (1-61^(1/2))/10

t_1 = (1-61^(1/2))/10

n^-1-((1-61^(1/2))/10) = 0

1*n^-1 = (1-61^(1/2))/10 // : 1

n^-1 = (1-61^(1/2))/10

-1 < 0

1/(n^1) = (1-61^(1/2))/10 // * n^1

1 = ((1-61^(1/2))/10)*n^1 // : (1-61^(1/2))/10

10*(1-61^(1/2))^-1 = n^1

n = 10*(1-61^(1/2))^-1

t_1 = (61^(1/2)+1)/10

n^-1-((61^(1/2)+1)/10) = 0

1*n^-1 = (61^(1/2)+1)/10 // : 1

n^-1 = (61^(1/2)+1)/10

-1 < 0

1/(n^1) = (61^(1/2)+1)/10 // * n^1

1 = ((61^(1/2)+1)/10)*n^1 // : (61^(1/2)+1)/10

10*(61^(1/2)+1)^-1 = n^1

n = 10*(61^(1/2)+1)^-1

n in { 10*(1-61^(1/2))^-1, 10*(61^(1/2)+1)^-1 }

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