3/2m+4=(1/m+2)-2

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Solution for 3/2m+4=(1/m+2)-2 equation: 


D( m )

m = 0

m = 0

m = 0

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

(3/2)*m+4 = 1/m-2+2 // - 1/m-2+2

(3/2)*m-(1/m)-2+2+4 = 0

(3/2)*m-m^-1-2+2+4 = 0

3/2*m^1-1*m^-1+4*m^0 = 0

(3/2*m^2+4*m^1-1*m^0)/(m^1) = 0 // * m^2

m^1*(3/2*m^2+4*m^1-1*m^0) = 0

m^1

(3/2)*m^2+4*m-1 = 0

(3/2)*m^2+4*m-1 = 0

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

DELTA = 22

DELTA > 0

m = (22^(1/2)-4)/(2*(3/2)) or m = (-22^(1/2)-4)/(2*(3/2))

m = (22^(1/2)-4)/3 or m = (-(22^(1/2)+4))/3

m in { (-(22^(1/2)+4))/3, (22^(1/2)-4)/3} 

m in { (-(22^(1/2)+4))/3, (22^(1/2)-4)/3 }

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