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

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

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

We move all terms to the left:

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


Domain of the equation: 3x+12)!=0

x∈R


We add all the numbers together, and all the variables

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

We get rid of parentheses

-1x-1/3x+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}
$