(10+3x^2+x^4)/(3-2x+x^2)

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Solution for (10+3x^2+x^4)/(3-2x+x^2) equation: 


D( x )

x^2-(2*x)+3 = 0

x^2-(2*x)+3 = 0

x^2-(2*x)+3 = 0

x^2-2*x+3 = 0

x^2-2*x+3 = 0

DELTA = (-2)^2-(1*3*4)

DELTA = -8

DELTA < 0

x in (-oo:+oo)

(x^4+3*x^2+10)/(x^2-(2*x)+3) = 0

(x^4+3*x^2+10)/(x^2-2*x+3) = 0

x^2-2*x+3 = 0

x^2-2*x+3 = 0

DELTA = (-2)^2-(1*3*4)

DELTA = -8

DELTA < 0

1 = 0

x^4+3*x^2+10 = 0

(x^4+3*x^2+10)/(x^2-(2*x)+3) = 0 // * x^2-(2*x)+3

x^4+3*x^2+10 = 0

t_1 = x^2

1*t_1^2+3*t_1^1+10 = 0

t_1^2+3*t_1+10 = 0

DELTA = 3^2-(1*4*10)

DELTA = -31

DELTA < 0

x belongs to the empty set

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