1/3X+12=x-4

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

1/3X+12=X-4

We move all terms to the left:

1/3X+12-(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+12=0

We multiply all the terms by the denominator


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

Wy multiply elements

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

We add all the numbers together, and all the variables

-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}
$