k-3/k+1=10/6

Solution for k-3/k+1=10/6 equation:

D( k )

k = 0

k = 0

k = 0

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

k-(3/k)+1 = 10/6 // - 10/6

k-(3/k)-(10/6)+1 = 0

k-3*k^-1-5/3+1 = 0

1*k^1-3*k^-1-2/3*k^0 = 0

(1*k^2-2/3*k^1-3*k^0)/(k^1) = 0 // * k^2

k^1*(1*k^2-2/3*k^1-3*k^0) = 0

k^1

k^2+(-2/3)*k-3 = 0

k^2+(-2/3)*k-3 = 0

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

DELTA = 112/9

DELTA > 0

k = ((112/9)^(1/2)-(-2/3))/(1*2) or k = (-(-2/3)-(112/9)^(1/2))/(1*2)

k = ((112/9)^(1/2)+2/3)/2 or k = (2/3-(112/9)^(1/2))/2

k in { (2/3-(112/9)^(1/2))/2, ((112/9)^(1/2)+2/3)/2} 

k in { (2/3-(112/9)^(1/2))/2, ((112/9)^(1/2)+2/3)/2 }