(8/3x)+(x-6/3)=2

Solution for (8/3x)+(x-6/3)=2 equation:

(8/3x)+(x-6/3)=2

We move all terms to the left:

(8/3x)+(x-6/3)-(2)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+8/3x)+(x-2)-2=0

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

3x^2-6x-6x+8=0

We add all the numbers together, and all the variables

3x^2-12x+8=0

a = 3; b = -12; c = +8;
Δ = b2-4ac
Δ = -122-4·3·8
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{3}}{2*3}=\frac{12-4\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*3}=\frac{12+4\sqrt{3}}{6}
$