m+4=2/3m-1

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

m+4=2/3m-1

We move all terms to the left:

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


Domain of the equation: 3m-1)!=0

m∈R


We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

3m^2+15m-2=0

a = 3; b = 15; c = -2;
Δ = b2-4ac
Δ = 152-4·3·(-2)
Δ = 249
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{-(15)-\sqrt{249}}{2*3}=\frac{-15-\sqrt{249}}{6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{249}}{2*3}=\frac{-15+\sqrt{249}}{6}
$