1/3x+4+12=x

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

1/3x+4+12=x

We move all terms to the left:

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


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

-1x+1/3x+16=0

We multiply all the terms by the denominator


-1x*3x+16*3x+1=0

Wy multiply elements

-3x^2+48x+1=0

a = -3; b = 48; c = +1;
Δ = b2-4ac
Δ = 482-4·(-3)·1
Δ = 2316
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:

$\sqrt{\Delta}=\sqrt{2316}=\sqrt{4*579}=\sqrt{4}*\sqrt{579}=2\sqrt{579}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-2\sqrt{579}}{2*-3}=\frac{-48-2\sqrt{579}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+2\sqrt{579}}{2*-3}=\frac{-48+2\sqrt{579}}{-6}
$