1/3x+4=-x+8

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

1/3x+4=-x+8

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

3x^2-12x+1=0

a = 3; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·3·1
Δ = 132
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:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{33}}{2*3}=\frac{12-2\sqrt{33}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{33}}{2*3}=\frac{12+2\sqrt{33}}{6}
$