1/3x+(x-4)=8

Solution for 1/3x+(x-4)=8 equation:

1/3x+(x-4)=8

We move all terms to the left:

1/3x+(x-4)-(8)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We get rid of parentheses

1/3x+x-4-8=0

We multiply all the terms by the denominator


x*3x-4*3x-8*3x+1=0

Wy multiply elements

3x^2-12x-24x+1=0

We add all the numbers together, and all the variables

3x^2-36x+1=0

a = 3; b = -36; c = +1;
Δ = b2-4ac
Δ = -362-4·3·1
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-2\sqrt{321}}{2*3}=\frac{36-2\sqrt{321}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+2\sqrt{321}}{2*3}=\frac{36+2\sqrt{321}}{6}
$