[x+1/3x+1/3x-4+2(1/3x-4)=72

Solution for [x+1/3x+1/3x-4+2(1/3x-4)=72 equation:

x+1/3x+1/3x-4+2(1/3x-4)=72

We move all terms to the left:

x+1/3x+1/3x-4+2(1/3x-4)-(72)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 3x-4)!=0

x∈R


We add all the numbers together, and all the variables

x+1/3x+1/3x+2(1/3x-4)-76=0

We multiply parentheses

x+1/3x+1/3x+2x-8-76=0

We multiply all the terms by the denominator


x*3x+2x*3x-8*3x-76*3x+1+1=0

We add all the numbers together, and all the variables

x*3x+2x*3x-8*3x-76*3x+2=0

Wy multiply elements

3x^2+6x^2-24x-228x+2=0

We add all the numbers together, and all the variables

9x^2-252x+2=0

a = 9; b = -252; c = +2;
Δ = b2-4ac
Δ = -2522-4·9·2
Δ = 63432
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{63432}=\sqrt{36*1762}=\sqrt{36}*\sqrt{1762}=6\sqrt{1762}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-252)-6\sqrt{1762}}{2*9}=\frac{252-6\sqrt{1762}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-252)+6\sqrt{1762}}{2*9}=\frac{252+6\sqrt{1762}}{18}
$