2/3m+8=m-10

Solution for 2/3m+8=m-10 equation:

2/3m+8=m-10

We move all terms to the left:

2/3m+8-(m-10)=0


Domain of the equation: 3m!=0

m!=0/3

m!=0

m∈R


We get rid of parentheses

2/3m-m+10+8=0

We multiply all the terms by the denominator


-m*3m+10*3m+8*3m+2=0

Wy multiply elements

-3m^2+30m+24m+2=0

We add all the numbers together, and all the variables

-3m^2+54m+2=0

a = -3; b = 54; c = +2;
Δ = b2-4ac
Δ = 542-4·(-3)·2
Δ = 2940
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:

$\sqrt{\Delta}=\sqrt{2940}=\sqrt{196*15}=\sqrt{196}*\sqrt{15}=14\sqrt{15}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-14\sqrt{15}}{2*-3}=\frac{-54-14\sqrt{15}}{-6}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+14\sqrt{15}}{2*-3}=\frac{-54+14\sqrt{15}}{-6}
$