-(4/5v-10)+3=-(1/v-2)

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Solution for -(4/5v-10)+3=-(1/v-2) equation:


D( v )

v = 0

v = 0

v = 0

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

3-((4/5)*v)+10 = -(1/v-2) // + 1/v-2

1/v-((4/5)*v)-2+3+10 = 0

(-4/5)*v+1/v-2+3+10 = 0

1*v^-1-4/5*v^1+11*v^0 = 0

(11*v^1-4/5*v^2+1*v^0)/(v^1) = 0 // * v^2

v^1*(11*v^1-4/5*v^2+1*v^0) = 0

v^1

(-4/5)*v^2+11*v+1 = 0

(-4/5)*v^2+11*v+1 = 0

DELTA = 11^2-(1*4*(-4/5))

DELTA = 621/5

DELTA > 0

v = ((621/5)^(1/2)-11)/(2*(-4/5)) or v = (-(621/5)^(1/2)-11)/(2*(-4/5))

v = -5/8*((621/5)^(1/2)-11) or v = 5/8*((621/5)^(1/2)+11)

v in { -5/8*((621/5)^(1/2)-11), 5/8*((621/5)^(1/2)+11)}

v in { -5/8*((621/5)^(1/2)-11), 5/8*((621/5)^(1/2)+11) }

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