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

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Solution for (2x^2-4x-4)/(5x+10)=0 equation:


D( x )

5*x+10 = 0

5*x+10 = 0

5*x+10 = 0

5*x+10 = 0 // - 10

5*x = -10 // : 5

x = -10/5

x = -2

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

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

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

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

2*(x^2-2*x-2) = 0

x^2-2*x-2 = 0

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

DELTA = 12

DELTA > 0

x = (12^(1/2)+2)/(1*2) or x = (2-12^(1/2))/(1*2)

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

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

(2*(x-((2-2*3^(1/2))/2))*(x-((2*3^(1/2)+2)/2)))/(5*x+10) = 0

( x-((2*3^(1/2)+2)/2) )

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

x = (2*3^(1/2)+2)/2

( x-((2-2*3^(1/2))/2) )

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

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

x in { (2*3^(1/2)+2)/2, (2-2*3^(1/2))/2 }

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