(1/3)m+0.1m+(1/12)m+0.20m+680=m

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Solution for (1/3)m+0.1m+(1/12)m+0.20m+680=m equation: 



(1/3)m+0.1m+(1/12)m+0.20m+680=m

We move all terms to the left:

(1/3)m+0.1m+(1/12)m+0.20m+680-(m)=0



Domain of the equation: 3)m!=0

m!=0/1

m!=0

m∈R





Domain of the equation: 12)m!=0

m!=0/1

m!=0

m∈R



We add all the numbers together, and all the variables

(+1/3)m+0.1m+(+1/12)m+0.20m-m+680=0

We add all the numbers together, and all the variables

-0.7m+(+1/3)m+(+1/12)m+680=0

We multiply parentheses

m^2+m^2-0.7m+680=0

We add all the numbers together, and all the variables

2m^2-0.7m+680=0

a = 2; b = -0.7; c = +680;
Δ = b2-4ac
Δ = -0.72-4·2·680
Δ = -5439.51
Delta is less than zero, so there is no solution for the equation

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