1/3m-4=m+2

Solution for 1/3m-4=m+2 equation:

1/3m-4=m+2

We move all terms to the left:

1/3m-4-(m+2)=0


Domain of the equation: 3m!=0

m!=0/3

m!=0

m∈R


We get rid of parentheses

1/3m-m-2-4=0

We multiply all the terms by the denominator


-m*3m-2*3m-4*3m+1=0

Wy multiply elements

-3m^2-6m-12m+1=0

We add all the numbers together, and all the variables

-3m^2-18m+1=0

a = -3; b = -18; c = +1;
Δ = b2-4ac
Δ = -182-4·(-3)·1
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-4\sqrt{21}}{2*-3}=\frac{18-4\sqrt{21}}{-6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+4\sqrt{21}}{2*-3}=\frac{18+4\sqrt{21}}{-6}
$