1/3a-12=4/4a+6

Solution for 1/3a-12=4/4a+6 equation:

1/3a-12=4/4a+6

We move all terms to the left:

1/3a-12-(4/4a+6)=0


Domain of the equation: 3a!=0

a!=0/3

a!=0

a∈R



Domain of the equation: 4a+6)!=0

a∈R


We get rid of parentheses

1/3a-4/4a-6-12=0

We calculate fractions


4a/12a^2+(-12a)/12a^2-6-12=0

We add all the numbers together, and all the variables

4a/12a^2+(-12a)/12a^2-18=0

We multiply all the terms by the denominator


4a+(-12a)-18*12a^2=0

Wy multiply elements

-216a^2+4a+(-12a)=0

We get rid of parentheses

-216a^2+4a-12a=0

We add all the numbers together, and all the variables

-216a^2-8a=0

a = -216; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-216)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-216}=\frac{0}{-432}
=0
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-216}=\frac{16}{-432}
=-1/27
$