((3)/(4x-8))+((x)/(5))=2

Solution for ((3)/(4x-8))+((x)/(5))=2 equation:

((3)/(4x-8))+((x)/(5))=2

We move all terms to the left:

((3)/(4x-8))+((x)/(5))-(2)=0


Domain of the equation: (4x-8))!=0

x∈R


We add all the numbers together, and all the variables

(3/(4x-8))+(+x/5)-2=0

We get rid of parentheses

(3/(4x-8))+x/5-2=0

We calculate fractions


(-4x^2)/20x+()/20x-2=0

We multiply all the terms by the denominator


(-4x^2)-2*20x+()=0

We add all the numbers together, and all the variables

(-4x^2)-2*20x=0

Wy multiply elements

(-4x^2)-40x=0

We get rid of parentheses

-4x^2-40x=0

a = -4; b = -40; c = 0;
Δ = b2-4ac
Δ = -402-4·(-4)·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40}{2*-4}=\frac{0}{-8}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40}{2*-4}=\frac{80}{-8}
=-10
$