6/(2x-4)=x-5/10

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Solution for 6/(2x-4)=x-5/10 equation: 


D( x )

2*x-4 = 0

2*x-4 = 0

2*x-4 = 0

2*x-4 = 0 // + 4

2*x = 4 // : 2

x = 4/2

x = 2

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

6/(2*x-4) = x-(5/10) // - x-(5/10)

6/(2*x-4)-x+5/10 = 0

6/(2*x-4)-x+1/2 = 0

(2*6)/(2*(2*x-4))+(2*(-x)*(2*x-4))/(2*(2*x-4))+(1*(2*x-4))/(2*(2*x-4)) = 0

2*(-x)*(2*x-4)+1*(2*x-4)+2*6 = 0

8*x-4*x^2+2*x-4+12 = 0

10*x-4*x^2+8 = 0

10*x-4*x^2+8 = 0

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

5*x-2*x^2+4 = 0

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

DELTA = 57

DELTA > 0

x = (57^(1/2)-5)/(-2*2) or x = (-57^(1/2)-5)/(-2*2)

x = (57^(1/2)-5)/(-4) or x = (57^(1/2)+5)/4

2*(x-((57^(1/2)-5)/(-4)))*(x-((57^(1/2)+5)/4)) = 0

(2*(x-((57^(1/2)-5)/(-4)))*(x-((57^(1/2)+5)/4)))/(2*(2*x-4)) = 0

(2*(x-((57^(1/2)-5)/(-4)))*(x-((57^(1/2)+5)/4)))/(2*(2*x-4)) = 0 // * 2*(2*x-4)

2*(x-((57^(1/2)-5)/(-4)))*(x-((57^(1/2)+5)/4)) = 0

( 2 )

2 = 0

x belongs to the empty set

( x-((57^(1/2)+5)/4) )

x-((57^(1/2)+5)/4) = 0 // + (57^(1/2)+5)/4

x = (57^(1/2)+5)/4

( x-((57^(1/2)-5)/(-4)) )

x-((57^(1/2)-5)/(-4)) = 0 // + (57^(1/2)-5)/(-4)

x = (57^(1/2)-5)/(-4)

x in { (57^(1/2)+5)/4, (57^(1/2)-5)/(-4) }

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