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

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

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

We move all terms to the left:

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


Domain of the equation: (2m-10)!=0

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

2m!=10

m!=10/2

m!=5

m∈R



Domain of the equation: 3m+10)!=0

m∈R


We get rid of parentheses

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

We calculate fractions


9m/(6m^2-30m)+(-4m+20)/(6m^2-30m)-10=0

We multiply all the terms by the denominator


9m+(-4m+20)-10*(6m^2-30m)=0

We multiply parentheses

-60m^2+9m+(-4m+20)+300m=0

We get rid of parentheses

-60m^2+9m-4m+300m+20=0

We add all the numbers together, and all the variables

-60m^2+305m+20=0

a = -60; b = 305; c = +20;
Δ = b2-4ac
Δ = 3052-4·(-60)·20
Δ = 97825
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{97825}=\sqrt{25*3913}=\sqrt{25}*\sqrt{3913}=5\sqrt{3913}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(305)-5\sqrt{3913}}{2*-60}=\frac{-305-5\sqrt{3913}}{-120}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(305)+5\sqrt{3913}}{2*-60}=\frac{-305+5\sqrt{3913}}{-120}
$