0.2+2k^2/3=0.1k

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Solution for 0.2+2k^2/3=0.1k equation: 


k in (-oo:+oo)

(2*k^2)/3+0.2 = 0.1*k // - 0.1*k

(2*k^2)/3-(0.1*k)+0.2 = 0

(2*k^2)/3-0.1*k+0.2 = 0

2/3*k^2-0.1*k+0.2 = 0

DELTA = (-0.1)^2-(0.2*2/3*4)

DELTA = 0.1^2-0.533333336

DELTA = -0.523333336

DELTA < 0

k belongs to the empty set

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