(w^2+19)/(w^2-3w-54)

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Solution for (w^2+19)/(w^2-3w-54) equation: 


D( w )

w^2-(3*w)-54 = 0

w^2-(3*w)-54 = 0

w^2-(3*w)-54 = 0

w^2-3*w-54 = 0

w^2-3*w-54 = 0

DELTA = (-3)^2-(-54*1*4)

DELTA = 225

DELTA > 0

w = (225^(1/2)+3)/(1*2) or w = (3-225^(1/2))/(1*2)

w = 9 or w = -6

w in (-oo:-6) U (-6:9) U (9:+oo)

(w^2+19)/(w^2-(3*w)-54) = 0

(w^2+19)/(w^2-3*w-54) = 0

w^2-3*w-54 = 0

w^2-3*w-54 = 0

DELTA = (-3)^2-(-54*1*4)

DELTA = 225

DELTA > 0

w = (225^(1/2)+3)/(1*2) or w = (3-225^(1/2))/(1*2)

w = 9 or w = -6

(w+6)*(w-9) = 0

(w^2+19)/((w+6)*(w-9)) = 0

1*w^2 = -19 // : 1

w^2 = -19

w belongs to the empty set

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