B=5/2(J-27)

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Solution for B=5/2(J-27) equation: 


x in (-oo:+oo)

B = (5/2)*(J-27) // - (5/2)*(J-27)

B-((5/2)*(J-27)) = 0

(-5/2)*(J-27)+B = 0

B-5/2*(J-27) = 0

B-5/2*(J-27) = 0

B-5/2*J+135/2 = 0

B-5/2*J+135/2 = 0

x belongs to the empty set

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