(-6/v+5)=(5/2v+10)+1

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Solution for (-6/v+5)=(5/2v+10)+1 equation: 


D( v )

v = 0

v = 0

v = 0

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

5-6/v = (5/2)*v+1+10 // - (5/2)*v+1+10

5-((5/2)*v)-6/v-10-1 = 0

(-5/2)*v-6/v-10-1+5 = 0

-5/2*v^1-6*v^-1-6*v^0 = 0

(-5/2*v^2-6*v^1-6*v^0)/(v^1) = 0 // * v^2

v^1*(-5/2*v^2-6*v^1-6*v^0) = 0

v^1

(-5/2)*v^2-6*v-6 = 0

(-5/2)*v^2-6*v-6 = 0

DELTA = (-6)^2-(-6*4*(-5/2))

DELTA = -24

DELTA < 0

v in { } 

v belongs to the empty set

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