1/n-4-2/n=3/4-n

Solution for 1/n-4-2/n=3/4-n equation:

D( n )

n = 0

n = 0

n = 0

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

1/n-(2/n)-4 = 3/4-n // - 3/4-n

n+1/n-(2/n)-(3/4)-4 = 0

n+1/n-2*n^-1-3/4-4 = 0

1*n^1-1*n^-1-19/4*n^0 = 0

(1*n^2-19/4*n^1-1*n^0)/(n^1) = 0 // * n^2

n^1*(1*n^2-19/4*n^1-1*n^0) = 0

n^1

n^2+(-19/4)*n-1 = 0

n^2+(-19/4)*n-1 = 0

DELTA = (-19/4)^2-(-1*1*4)

DELTA = 425/16

DELTA > 0

n = ((425/16)^(1/2)-(-19/4))/(1*2) or n = (-(-19/4)-(425/16)^(1/2))/(1*2)

n = ((425/16)^(1/2)+19/4)/2 or n = (19/4-(425/16)^(1/2))/2

n in { (19/4-(425/16)^(1/2))/2, ((425/16)^(1/2)+19/4)/2} 

n in { (19/4-(425/16)^(1/2))/2, ((425/16)^(1/2)+19/4)/2 }