3/(m+4)-4/m=6

Solution for 3/(m+4)-4/m=6 equation:

3/(m+4)-4/m=6

We move all terms to the left:

3/(m+4)-4/m-(6)=0


Domain of the equation: (m+4)!=0

We move all terms containing m to the left, all other terms to the right

m!=-4

m∈R



Domain of the equation: m!=0

m∈R


We calculate fractions


3m/(m^2+4m)+(-4m-16)/(m^2+4m)-6=0

We multiply all the terms by the denominator


3m+(-4m-16)-6*(m^2+4m)=0

We multiply parentheses

-6m^2+3m+(-4m-16)-24m=0

We get rid of parentheses

-6m^2+3m-4m-24m-16=0

We add all the numbers together, and all the variables

-6m^2-25m-16=0

a = -6; b = -25; c = -16;
Δ = b2-4ac
Δ = -252-4·(-6)·(-16)
Δ = 241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-\sqrt{241}}{2*-6}=\frac{25-\sqrt{241}}{-12}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{241}}{2*-6}=\frac{25+\sqrt{241}}{-12}
$