-2/3x-4+x=2

Solution for -2/3x-4+x=2 equation:

-2/3x-4+x=2

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

x-2/3x-6=0

We multiply all the terms by the denominator


x*3x-6*3x-2=0

Wy multiply elements

3x^2-18x-2=0

a = 3; b = -18; c = -2;
Δ = b2-4ac
Δ = -182-4·3·(-2)
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{87}}{2*3}=\frac{18-2\sqrt{87}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{87}}{2*3}=\frac{18+2\sqrt{87}}{6}
$