1/3x+12=9x+8

Solution for 1/3x+12=9x+8 equation:

1/3x+12=9x+8

We move all terms to the left:

1/3x+12-(9x+8)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We get rid of parentheses

1/3x-9x-8+12=0

We multiply all the terms by the denominator


-9x*3x-8*3x+12*3x+1=0

Wy multiply elements

-27x^2-24x+36x+1=0

We add all the numbers together, and all the variables

-27x^2+12x+1=0

a = -27; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-27)·1
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{7}}{2*-27}=\frac{-12-6\sqrt{7}}{-54}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{7}}{2*-27}=\frac{-12+6\sqrt{7}}{-54}
$