1/3x-4(2/3x+3)=2/3x-6

Solution for 1/3x-4(2/3x+3)=2/3x-6 equation:

1/3x-4(2/3x+3)=2/3x-6

We move all terms to the left:

1/3x-4(2/3x+3)-(2/3x-6)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 3x+3)!=0

x∈R



Domain of the equation: 3x-6)!=0

x∈R


We multiply parentheses

1/3x-8x-(2/3x-6)-12=0

We get rid of parentheses

1/3x-8x-2/3x+6-12=0

We multiply all the terms by the denominator


-8x*3x+6*3x-12*3x+1-2=0

We add all the numbers together, and all the variables

-8x*3x+6*3x-12*3x-1=0

Wy multiply elements

-24x^2+18x-36x-1=0

We add all the numbers together, and all the variables

-24x^2-18x-1=0

a = -24; b = -18; c = -1;
Δ = b2-4ac
Δ = -182-4·(-24)·(-1)
Δ = 228
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:

$\sqrt{\Delta}=\sqrt{228}=\sqrt{4*57}=\sqrt{4}*\sqrt{57}=2\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{57}}{2*-24}=\frac{18-2\sqrt{57}}{-48}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{57}}{2*-24}=\frac{18+2\sqrt{57}}{-48}
$