1=20y-(5y/y^2)

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Solution for 1=20y-(5y/y^2) equation:


D( y )

y^2 = 0

y^2 = 0

y^2 = 0

1*y^2 = 0 // : 1

y^2 = 0

y = 0

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

1 = 20*y-((5*y)/(y^2)) // - 20*y-((5*y)/(y^2))

(5*y)/(y^2)-(20*y)+1 = 0

(5*y)/(y^2)-20*y+1 = 0

5*y^-1-20*y^1+1*y^0 = 0

(1*y^1-20*y^2+5*y^0)/(y^1) = 0 // * y^2

y^1*(1*y^1-20*y^2+5*y^0) = 0

y^1

y-20*y^2+5 = 0

y-20*y^2+5 = 0

DELTA = 1^2-(-20*4*5)

DELTA = 401

DELTA > 0

y = (401^(1/2)-1)/(-20*2) or y = (-401^(1/2)-1)/(-20*2)

y = (401^(1/2)-1)/(-40) or y = (401^(1/2)+1)/40

y in { (401^(1/2)-1)/(-40), (401^(1/2)+1)/40}

y in { (401^(1/2)-1)/(-40), (401^(1/2)+1)/40 }

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