-3/4x+1=4/3x+12

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

-3/4x+1=4/3x+12

We move all terms to the left:

-3/4x+1-(4/3x+12)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 3x+12)!=0

x∈R


We get rid of parentheses

-3/4x-4/3x-12+1=0

We calculate fractions


(-9x)/12x^2+(-16x)/12x^2-12+1=0

We add all the numbers together, and all the variables

(-9x)/12x^2+(-16x)/12x^2-11=0

We multiply all the terms by the denominator


(-9x)+(-16x)-11*12x^2=0

Wy multiply elements

-132x^2+(-9x)+(-16x)=0

We get rid of parentheses

-132x^2-9x-16x=0

We add all the numbers together, and all the variables

-132x^2-25x=0

a = -132; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·(-132)·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*-132}=\frac{0}{-264}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*-132}=\frac{50}{-264}
=-25/132
$