(1+z+z^2+8)/(z+1+8-z+z^2-7z-8)

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Solution for (1+z+z^2+8)/(z+1+8-z+z^2-7z-8) equation: 


D( z )

z^2+z-z-(7*z)-8+1+8 = 0

z^2+z-z-(7*z)-8+1+8 = 0

z^2+z-z-(7*z)-8+1+8 = 0

z^2+z-z-7*z-8+1+8 = 0

z^2-7*z+1 = 0

DELTA = (-7)^2-(1*1*4)

DELTA = 45

DELTA > 0

z = (45^(1/2)+7)/(1*2) or z = (7-45^(1/2))/(1*2)

z = (3*5^(1/2)+7)/2 or z = (7-3*5^(1/2))/2

z in (-oo:(7-3*5^(1/2))/2) U ((7-3*5^(1/2))/2:(3*5^(1/2)+7)/2) U ((3*5^(1/2)+7)/2:+oo)

(z^2+z+1+8)/(z^2+z-z-(7*z)-8+1+8) = 0

(z^2+z+1+8)/(z^2+z-z-7*z-8+1+8) = 0

(z^2+z+9)/(z^2-7*z+1) = 0

z^2+z+9 = 0

z^2+z+9 = 0

DELTA = 1^2-(1*4*9)

DELTA = -35

DELTA < 0

1 = 0

z^2-7*z+1 = 0

z^2-7*z+1 = 0

DELTA = (-7)^2-(1*1*4)

DELTA = 45

DELTA > 0

z = (45^(1/2)+7)/(1*2) or z = (7-45^(1/2))/(1*2)

z = (3*5^(1/2)+7)/2 or z = (7-3*5^(1/2))/2

(z-((7-3*5^(1/2))/2))*(z-((3*5^(1/2)+7)/2)) = 0

1/((z-((7-3*5^(1/2))/2))*(z-((3*5^(1/2)+7)/2))) = 0

z belongs to the empty set

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