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

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Solution for 2(1/3k-3)^2+5/3k-3+2=0 equation: 


k in (-oo:+oo)

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

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

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

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

2/3*k^2-7*k-9+6+54 = 0

2/3*k^2-7*k+6+45 = 0

2/3*k^2-7*k+51 = 0

2/3*k^2-7*k+51 = 0

2/3*k^2-7*k+51 = 0

DELTA = (-7)^2-(2/3*4*51)

DELTA = -87

DELTA < 0

1 = 0

1/3 = 0

1/3 = 0 // * 3

1 = 0

k belongs to the empty set

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