(10/m)+(3/2)=(m/10)

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Solution for (10/m)+(3/2)=(m/10) equation: 


D( m )

m = 0

m = 0

m = 0

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

10/m+3/2 = m/10 // - m/10

10/m-(m/10)+3/2 = 0

10*m^-1-1/10*m^1+3/2*m^0 = 0

(3/2*m^1-1/10*m^2+10*m^0)/(m^1) = 0 // * m^2

m^1*(3/2*m^1-1/10*m^2+10*m^0) = 0

m^1

(-1/10)*m^2+(3/2)*m+10 = 0

(-1/10)*m^2+(3/2)*m+10 = 0

DELTA = (3/2)^2-(4*10*(-1/10))

DELTA = 25/4

DELTA > 0

m = ((25/4)^(1/2)-(3/2))/(2*(-1/10)) or m = (-(3/2)-(25/4)^(1/2))/(2*(-1/10))

m = -5 or m = 20

m in { -5, 20} 

m in { -5, 20 }

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