8/4x+12=9/3x+20

Solution for 8/4x+12=9/3x+20 equation:

8/4x+12=9/3x+20

We move all terms to the left:

8/4x+12-(9/3x+20)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 3x+20)!=0

x∈R


We get rid of parentheses

8/4x-9/3x-20+12=0

We calculate fractions


24x/12x^2+(-36x)/12x^2-20+12=0

We add all the numbers together, and all the variables

24x/12x^2+(-36x)/12x^2-8=0

We multiply all the terms by the denominator


24x+(-36x)-8*12x^2=0

Wy multiply elements

-96x^2+24x+(-36x)=0

We get rid of parentheses

-96x^2+24x-36x=0

We add all the numbers together, and all the variables

-96x^2-12x=0

a = -96; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·(-96)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*-96}=\frac{0}{-192}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*-96}=\frac{24}{-192}
=-1/8
$