(2-2x^2+8x)/(1+x^2)^2=0

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Solution for (2-2x^2+8x)/(1+x^2)^2=0 equation: 


D( x )

(x^2+1)^2 = 0

(x^2+1)^2 = 0

(x^2+1)^2 = 0

1*x^2 = -1 // : 1

x^2 = -1

x in (-oo:+oo)

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

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

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

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

4*x-x^2+1 = 0

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

DELTA = 20

DELTA > 0

x = (20^(1/2)-4)/(-1*2) or x = (-20^(1/2)-4)/(-1*2)

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

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

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

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

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

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

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

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

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

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

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