(x+3)/(2x^2+9x-5)/(x^2+3x)/(2x-1)

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Solution for (x+3)/(2x^2+9x-5)/(x^2+3x)/(2x-1) equation: 


D( x )

x^2+3*x = 0

2*x-1 = 0

2*x^2+9*x-5 = 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

2*x-1 = 0

2*x-1 = 0

2*x-1 = 0 // + 1

2*x = 1 // : 2

x = 1/2

2*x^2+9*x-5 = 0

2*x^2+9*x-5 = 0

2*x^2+9*x-5 = 0

DELTA = 9^2-(-5*2*4)

DELTA = 121

DELTA > 0

x = (121^(1/2)-9)/(2*2) or x = (-121^(1/2)-9)/(2*2)

x = 1/2 or x = -5

x in (-oo:-5) U (-5:-3) U (-3:0) U (0:1/2) U (1/2:+oo)

(((x+3)/(2*x^2+9*x-5))/(x^2+3*x))/(2*x-1) = 0

(x+3)/((2*x^2+9*x-5)*(x^2+3*x)*(2*x-1)) = 0

2*x^2+9*x-5 = 0

2*x^2+9*x-5 = 0

DELTA = 9^2-(-5*2*4)

DELTA = 121

DELTA > 0

x = (121^(1/2)-9)/(2*2) or x = (-121^(1/2)-9)/(2*2)

x = 1/2 or x = -5

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

x^2+3*x = 0

x*(x+3) = 0

x+3 = 0 // - 3

x = -3

x*(x+3) = 0

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

x+3 = 0 // - 3

x = -3

x in { -3} 

x belongs to the empty set

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