(k-3)/(2k)=(3/6k)-1/4

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Solution for (k-3)/(2k)=(3/6k)-1/4 equation: 


D( k )

2*k = 0

2*k = 0

2*k = 0

2*k = 0 // : 2

k = 0

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

(k-3)/(2*k) = (3/6)*k-(1/4) // - (3/6)*k-(1/4)

(k-3)/(2*k)-((3/6)*k)+1/4 = 0

(k-3)/(2*k)+(-1/2)*k+1/4 = 0

(k-3)/(2*k)-k/2+1/4 = 0

(2*4*(k-3))/(2*2*4*k)+(2*4*(-k)*k)/(2*2*4*k)+(1*2*2*k)/(2*2*4*k) = 0

2*4*(k-3)+2*4*(-k)*k+1*2*2*k = 0

8*k-8*k^2+4*k-24 = 0

12*k-8*k^2-24 = 0

12*k-8*k^2-24 = 0

4*(3*k-2*k^2-6) = 0

3*k-2*k^2-6 = 0

DELTA = 3^2-(-6*(-2)*4)

DELTA = -39

DELTA < 0

4 = 0

4/(2*2*4*k) = 0

4/(2*2*4*k) = 0 // * 2*2*4*k

4 = 0

k belongs to the empty set

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