(7/2y-12)+1=(-4/y-6)

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Solution for (7/2y-12)+1=(-4/y-6) equation: 


D( y )

y = 0

y = 0

y = 0

y in (-oo:0) U (0:+oo)

(7/2)*y-12+1 = -4/y-6 // + -4/y-6

(7/2)*y-(-4/y)-12+1+6 = 0

(7/2)*y+4*y^-1-12+1+6 = 0

7/2*y^1+4*y^-1-5*y^0 = 0

(7/2*y^2-5*y^1+4*y^0)/(y^1) = 0 // * y^2

y^1*(7/2*y^2-5*y^1+4*y^0) = 0

y^1

(7/2)*y^2-5*y+4 = 0

(7/2)*y^2-5*y+4 = 0

DELTA = (-5)^2-(4*4*(7/2))

DELTA = -31

DELTA < 0

y in { } 

y belongs to the empty set

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