1/3x+8=x+4

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

1/3x+8=x+4

We move all terms to the left:

1/3x+8-(x+4)=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+12x+1=0

a = -3; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-3)·1
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{39}}{2*-3}=\frac{-12-2\sqrt{39}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{39}}{2*-3}=\frac{-12+2\sqrt{39}}{-6}
$