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

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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 $

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