12/(x^2+3x)+7/x+4/(x+3)

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Solution for 12/(x^2+3x)+7/x+4/(x+3) equation: 


D( x )

x^2+3*x = 0

x = 0

x+3 = 0

x^2+3*x = 0

x^2+3*x = 0

x^2+3*x = 0

DELTA = 3^2-(0*1*4)

DELTA = 9

DELTA > 0

x = (9^(1/2)-3)/(1*2) or x = (-9^(1/2)-3)/(1*2)

x = 0 or x = -3

x = 0

x = 0

x+3 = 0

x+3 = 0

x+3 = 0 // - 3

x = -3

x in (-oo:-3) U (-3:0) U (0:+oo)

12/(x^2+3*x)+4/(x+3)+7/x = 0

x^2+3*x = 0

x^2+3*x = 0

x*(x+3) = 0

x+3 = 0 // - 3

x = -3

x*(x+3) = 0

12/(x*(x+3))+4/(x+3)+7/x = 0

12/(x*(x+3))+(4*x)/(x*(x+3))+(7*(x+3))/(x*(x+3)) = 0

7*(x+3)+4*x+12 = 0

4*x+7*x+12+21 = 0

11*x+33 = 0

(11*x+33)/(x*(x+3)) = 0

(11*x+33)/(x*(x+3)) = 0 // * x*(x+3)

11*x+33 = 0

11*x+33 = 0 // - 33

11*x = -33 // : 11

x = -33/11

x = -3

x in { -3} 

x belongs to the empty set

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