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

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

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

We move all terms to the left:

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


Domain of the equation: 3)x!=0

x!=0/1

x!=0

x∈R



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

x∈R


We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses

x^2+2x+2x-8-76=0

We add all the numbers together, and all the variables

x^2+4x-84=0

a = 1; b = 4; c = -84;
Δ = b2-4ac
Δ = 42-4·1·(-84)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{22}}{2*1}=\frac{-4-4\sqrt{22}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{22}}{2*1}=\frac{-4+4\sqrt{22}}{2}
$