1/3x+(12-x)=39

Solution for 1/3x+(12-x)=39 equation:

1/3x+(12-x)=39

We move all terms to the left:

1/3x+(12-x)-(39)=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+12)-39=0

We get rid of parentheses

1/3x-1x+12-39=0

We multiply all the terms by the denominator


-1x*3x+12*3x-39*3x+1=0

Wy multiply elements

-3x^2+36x-117x+1=0

We add all the numbers together, and all the variables

-3x^2-81x+1=0

a = -3; b = -81; c = +1;
Δ = b2-4ac
Δ = -812-4·(-3)·1
Δ = 6573
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-81)-\sqrt{6573}}{2*-3}=\frac{81-\sqrt{6573}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-81)+\sqrt{6573}}{2*-3}=\frac{81+\sqrt{6573}}{-6}
$