(m+4)/m+m/3=m/3

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

(m+4)/m+m/3=m/3

We move all terms to the left:

(m+4)/m+m/3-(m/3)=0


Domain of the equation: m!=0

m∈R


We add all the numbers together, and all the variables

(m+4)/m+m/3-(+m/3)=0

We get rid of parentheses

(m+4)/m+m/3-m/3=0

We calculate fractions


(-1m^2+m)/3m+(3m+12)/3m=0

We multiply all the terms by the denominator


(-1m^2+m)+(3m+12)=0

We get rid of parentheses

-1m^2+m+3m+12=0

We add all the numbers together, and all the variables

-1m^2+4m+12=0

a = -1; b = 4; c = +12;
Δ = b2-4ac
Δ = 42-4·(-1)·12
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*-1}=\frac{-12}{-2}
=+6
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*-1}=\frac{4}{-2}
=-2
$