3/(x-3)=2/(x-3)+8/(x^2-7+12)

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Solution for 3/(x-3)=2/(x-3)+8/(x^2-7+12) equation: 


D( x )

x^2-7+12 = 0

x-3 = 0

x^2-7+12 = 0

x^2-7+12 = 0

1*x^2 = -5 // : 1

x^2 = -5

x-3 = 0

x-3 = 0

x-3 = 0 // + 3

x = 3

x in (-oo:3) U (3:+oo)

3/(x-3) = 2/(x-3)+8/(x^2-7+12) // - 2/(x-3)+8/(x^2-7+12)

3/(x-3)-(2/(x-3))-(8/(x^2-7+12)) = 0

3/(x-3)-2*(x-3)^-1-8*(x^2-7+12)^-1 = 0

3/(x-3)-2/(x-3)-8/(x^2+5) = 0

(3*(x^2+5))/((x-3)*(x^2+5))+(-2*(x^2+5))/((x-3)*(x^2+5))+(-8*(x-3))/((x-3)*(x^2+5)) = 0

3*(x^2+5)-2*(x^2+5)-8*(x-3) = 0

x^2-8*x+5+24 = 0

x^2-8*x+29 = 0

x^2-8*x+29 = 0

x^2-8*x+29 = 0

DELTA = (-8)^2-(1*4*29)

DELTA = -52

DELTA < 0

1 = 0

1/((x-3)*(x^2+5)) = 0

1/((x-3)*(x^2+5)) = 0 // * (x-3)*(x^2+5)

1 = 0

x belongs to the empty set

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