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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)!=0We get rid of parentheses
m∈R
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} $
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