4b+12/b+3*b-5/4

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Solution for 4b+12/b+3*b-5/4 equation: 


D( b )

b = 0

b = 0

b = 0

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

4*b+3*b+12/b-(5/4) = 0

4*b+3*b+12/b-5/4 = 0

7*b^1+12*b^-1-5/4*b^0 = 0

(7*b^2-5/4*b^1+12*b^0)/(b^1) = 0 // * b^2

b^1*(7*b^2-5/4*b^1+12*b^0) = 0

b^1

7*b^2+(-5/4)*b+12 = 0

7*b^2+(-5/4)*b+12 = 0

DELTA = (-5/4)^2-(4*7*12)

DELTA = -5351/16

DELTA < 0

b in { } 

b belongs to the empty set

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