1/3x+12x=x-4

Solution for 1/3x+12x=x-4 equation:

1/3x+12x=x-4

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

12x+1/3x-x+4=0

We multiply all the terms by the denominator


12x*3x-x*3x+4*3x+1=0

Wy multiply elements

36x^2-3x^2+12x+1=0

We add all the numbers together, and all the variables

33x^2+12x+1=0

a = 33; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·33·1
Δ = 12
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:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{3}}{2*33}=\frac{-12-2\sqrt{3}}{66}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{3}}{2*33}=\frac{-12+2\sqrt{3}}{66}
$