((3)/(5x+2))+((5x)/(25x2-4))

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Solution for ((3)/(5x+2))+((5x)/(25x2-4)) equation: 


D( x )

25*x^2-4 = 0

5*x+2 = 0

25*x^2-4 = 0

25*x^2-4 = 0

25*x^2 = 4 // : 25

x^2 = 4/25

x^2 = 4/25 // ^ 1/2

abs(x) = 2/5

x = 2/5 or x = -2/5

5*x+2 = 0

5*x+2 = 0

5*x+2 = 0 // - 2

5*x = -2 // : 5

x = -2/5

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

3/(5*x+2)+(5*x)/(25*x^2-4) = 0

(3*(25*x^2-4))/((5*x+2)*(25*x^2-4))+(5*x*(5*x+2))/((5*x+2)*(25*x^2-4)) = 0

3*(25*x^2-4)+5*x*(5*x+2) = 0

100*x^2+10*x-12 = 0

100*x^2+10*x-12 = 0

2*(50*x^2+5*x-6) = 0

50*x^2+5*x-6 = 0

DELTA = 5^2-(-6*4*50)

DELTA = 1225

DELTA > 0

x = (1225^(1/2)-5)/(2*50) or x = (-1225^(1/2)-5)/(2*50)

x = 3/10 or x = -2/5

2*(x+2/5)*(x-3/10) = 0

(2*(x+2/5)*(x-3/10))/((5*x+2)*(25*x^2-4)) = 0

(2*(x+2/5)*(x-3/10))/((5*x+2)*(25*x^2-4)) = 0 // * (5*x+2)*(25*x^2-4)

2*(x+2/5)*(x-3/10) = 0

( x+2/5 )

x+2/5 = 0 // - 2/5

x = -2/5

( x-3/10 )

x-3/10 = 0 // + 3/10

x = 3/10

x in { -2/5} 

x = 3/10

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