(-7/5v+10)+1=(3/v+2)

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Solution for (-7/5v+10)+1=(3/v+2) equation: 


D( v )

v = 0

v = 0

v = 0

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

(-7/5)*v+1+10 = 3/v+2 // - 3/v+2

(-7/5)*v-(3/v)-2+1+10 = 0

(-7/5)*v-3*v^-1-2+1+10 = 0

9*v^0-7/5*v^1-3*v^-1 = 0

(9*v^1-7/5*v^2-3*v^0)/(v^1) = 0 // * v^2

v^1*(9*v^1-7/5*v^2-3*v^0) = 0

v^1

(-7/5)*v^2+9*v-3 = 0

(-7/5)*v^2+9*v-3 = 0

DELTA = 9^2-(-3*4*(-7/5))

DELTA = 321/5

DELTA > 0

v = ((321/5)^(1/2)-9)/(2*(-7/5)) or v = (-(321/5)^(1/2)-9)/(2*(-7/5))

v = -5/14*((321/5)^(1/2)-9) or v = 5/14*((321/5)^(1/2)+9)

v in { -5/14*((321/5)^(1/2)-9), 5/14*((321/5)^(1/2)+9)} 

v in { -5/14*((321/5)^(1/2)-9), 5/14*((321/5)^(1/2)+9) }

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