(x-6/5x)+1=(x+2/x)

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


D( x )

x = 0

x = 0

x = 0

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

x-((6/5)*x)+1 = x+2/x // - x+2/x

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

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

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

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

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

x^1

(-6/5)*x^2+x-2 = 0

(-6/5)*x^2+x-2 = 0

DELTA = 1^2-(-2*4*(-6/5))

DELTA = -43/5

DELTA < 0

x in { } 

x belongs to the empty set

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