2/3m+5=21/6m-18

Solution for 2/3m+5=21/6m-18 equation:

2/3m+5=21/6m-18

We move all terms to the left:

2/3m+5-(21/6m-18)=0


Domain of the equation: 3m!=0

m!=0/3

m!=0

m∈R



Domain of the equation: 6m-18)!=0

m∈R


We get rid of parentheses

2/3m-21/6m+18+5=0

We calculate fractions


12m/18m^2+(-63m)/18m^2+18+5=0

We add all the numbers together, and all the variables

12m/18m^2+(-63m)/18m^2+23=0

We multiply all the terms by the denominator


12m+(-63m)+23*18m^2=0

Wy multiply elements

414m^2+12m+(-63m)=0

We get rid of parentheses

414m^2+12m-63m=0

We add all the numbers together, and all the variables

414m^2-51m=0

a = 414; b = -51; c = 0;
Δ = b2-4ac
Δ = -512-4·414·0
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2601}=51$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-51)-51}{2*414}=\frac{0}{828}
=0
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-51)+51}{2*414}=\frac{102}{828}
=17/138
$