(3x+12)/(x^2+3x)/(x^2+x-12)/(x^2-9x)

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Solution for (3x+12)/(x^2+3x)/(x^2+x-12)/(x^2-9x) equation: 


D( x )

x^2+3*x = 0

x^2+x-12 = 0

x^2-(9*x) = 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^2+x-12 = 0

x^2+x-12 = 0

x^2+x-12 = 0

DELTA = 1^2-(-12*1*4)

DELTA = 49

DELTA > 0

x = (49^(1/2)-1)/(1*2) or x = (-49^(1/2)-1)/(1*2)

x = 3 or x = -4

x^2-(9*x) = 0

x^2-(9*x) = 0

x^2-9*x = 0

x^2-9*x = 0

DELTA = (-9)^2-(0*1*4)

DELTA = 81

DELTA > 0

x = (81^(1/2)+9)/(1*2) or x = (9-81^(1/2))/(1*2)

x = 9 or x = 0

x in (-oo:-4) U (-4:-3) U (-3:0) U (0:3) U (3:9) U (9:+oo)

(((3*x+12)/(x^2+3*x))/(x^2+x-12))/(x^2-(9*x)) = 0

(((3*x+12)/(x^2+3*x))/(x^2+x-12))/(x^2-9*x) = 0

(3*x+12)/((x^2+3*x)*(x^2+x-12)*(x^2-9*x)) = 0

x^2+3*x = 0

x*(x+3) = 0

x+3 = 0 // - 3

x = -3

x*(x+3) = 0

x^2+x-12 = 0

x^2+x-12 = 0

DELTA = 1^2-(-12*1*4)

DELTA = 49

DELTA > 0

x = (49^(1/2)-1)/(1*2) or x = (-49^(1/2)-1)/(1*2)

x = 3 or x = -4

(x+4)*(x-3) = 0

x^2-9*x = 0

x*(x-9) = 0

x-9 = 0 // + 9

x = 9

x*(x-9) = 0

(3*x+12)/(x^2*(x+3)*(x+4)*(x-3)*(x-9)) = 0

3*x+12 = 0 // - 12

3*x = -12 // : 3

x = -12/3

x = -4

x in { -4} 

x belongs to the empty set

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