1/3x+12=180-x

Solution for 1/3x+12=180-x equation:

1/3x+12=180-x

We move all terms to the left:

1/3x+12-(180-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

1/3x-(-1x+180)+12=0

We get rid of parentheses

1/3x+1x-180+12=0

We multiply all the terms by the denominator


1x*3x-180*3x+12*3x+1=0

Wy multiply elements

3x^2-540x+36x+1=0

We add all the numbers together, and all the variables

3x^2-504x+1=0

a = 3; b = -504; c = +1;
Δ = b2-4ac
Δ = -5042-4·3·1
Δ = 254004
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:

$\sqrt{\Delta}=\sqrt{254004}=\sqrt{4*63501}=\sqrt{4}*\sqrt{63501}=2\sqrt{63501}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-504)-2\sqrt{63501}}{2*3}=\frac{504-2\sqrt{63501}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-504)+2\sqrt{63501}}{2*3}=\frac{504+2\sqrt{63501}}{6}
$