(x^2+4x+4)/(4x+8)

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Solution for (x^2+4x+4)/(4x+8) equation: 


D( x )

4*x+8 = 0

4*x+8 = 0

4*x+8 = 0

4*x+8 = 0 // - 8

4*x = -8 // : 4

x = -8/4

x = -2

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

(x^2+4*x+4)/(4*x+8) = 0

x^2+4*x+4 = 0

x^2+4*x+4 = 0

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

DELTA = 0

x = -4/(1*2)

x = -2 or x = -2

(x+2)^2 = 0

((x+2)^2)/(4*x+8) = 0

x+2 = 0 // - 2

x = -2

x in { -2} 

x belongs to the empty set

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