1/3x+(12-x)=38

Solution for 1/3x+(12-x)=38 equation:

1/3x+(12-x)=38

We move all terms to the left:

1/3x+(12-x)-(38)=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)-38=0

We get rid of parentheses

1/3x-1x+12-38=0

We multiply all the terms by the denominator


-1x*3x+12*3x-38*3x+1=0

Wy multiply elements

-3x^2+36x-114x+1=0

We add all the numbers together, and all the variables

-3x^2-78x+1=0

a = -3; b = -78; c = +1;
Δ = b2-4ac
Δ = -782-4·(-3)·1
Δ = 6096
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:

$\sqrt{\Delta}=\sqrt{6096}=\sqrt{16*381}=\sqrt{16}*\sqrt{381}=4\sqrt{381}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-78)-4\sqrt{381}}{2*-3}=\frac{78-4\sqrt{381}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-78)+4\sqrt{381}}{2*-3}=\frac{78+4\sqrt{381}}{-6}
$