(b/4)-(3/b)=(7b+5/28)

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Solution for (b/4)-(3/b)=(7b+5/28) equation: 


D( b )

b = 0

b = 0

b = 0

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

b/4-(3/b) = 7*b+5/28 // - 7*b+5/28

b/4-(7*b)-(3/b)-(5/28) = 0

b/4-7*b-3*b^-1-5/28 = 0

-27/4*b^1-3*b^-1-5/28*b^0 = 0

(-27/4*b^2-5/28*b^1-3*b^0)/(b^1) = 0 // * b^2

b^1*(-27/4*b^2-5/28*b^1-3*b^0) = 0

b^1

(-27/4)*b^2+(-5/28)*b-3 = 0

(-27/4)*b^2+(-5/28)*b-3 = 0

DELTA = (-5/28)^2-(-3*4*(-27/4))

DELTA = -63479/784

DELTA < 0

b in { } 

b belongs to the empty set

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