1/3x+12=3x+24

Solution for 1/3x+12=3x+24 equation:

1/3x+12=3x+24

We move all terms to the left:

1/3x+12-(3x+24)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We get rid of parentheses

1/3x-3x-24+12=0

We multiply all the terms by the denominator


-3x*3x-24*3x+12*3x+1=0

Wy multiply elements

-9x^2-72x+36x+1=0

We add all the numbers together, and all the variables

-9x^2-36x+1=0

a = -9; b = -36; c = +1;
Δ = b2-4ac
Δ = -362-4·(-9)·1
Δ = 1332
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:

$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-6\sqrt{37}}{2*-9}=\frac{36-6\sqrt{37}}{-18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+6\sqrt{37}}{2*-9}=\frac{36+6\sqrt{37}}{-18}
$