2/3x-4=3/4x+12

Solution for 2/3x-4=3/4x+12 equation:

2/3x-4=3/4x+12

We move all terms to the left:

2/3x-4-(3/4x+12)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R



Domain of the equation: 4x+12)!=0

x∈R


We get rid of parentheses

2/3x-3/4x-12-4=0

We calculate fractions


8x/12x^2+(-9x)/12x^2-12-4=0

We add all the numbers together, and all the variables

8x/12x^2+(-9x)/12x^2-16=0

We multiply all the terms by the denominator


8x+(-9x)-16*12x^2=0

Wy multiply elements

-192x^2+8x+(-9x)=0

We get rid of parentheses

-192x^2+8x-9x=0

We add all the numbers together, and all the variables

-192x^2-1x=0

a = -192; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-192)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-192}=\frac{0}{-384}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-192}=\frac{2}{-384}
=-1/192
$