42/4x+2=39/5x-2

Solution for 42/4x+2=39/5x-2 equation:

42/4x+2=39/5x-2

We move all terms to the left:

42/4x+2-(39/5x-2)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R



Domain of the equation: 5x-2)!=0

x∈R


We get rid of parentheses

42/4x-39/5x+2+2=0

We calculate fractions


210x/20x^2+(-156x)/20x^2+2+2=0

We add all the numbers together, and all the variables

210x/20x^2+(-156x)/20x^2+4=0

We multiply all the terms by the denominator


210x+(-156x)+4*20x^2=0

Wy multiply elements

80x^2+210x+(-156x)=0

We get rid of parentheses

80x^2+210x-156x=0

We add all the numbers together, and all the variables

80x^2+54x=0

a = 80; b = 54; c = 0;
Δ = b2-4ac
Δ = 542-4·80·0
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-54}{2*80}=\frac{-108}{160}
=-27/40
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+54}{2*80}=\frac{0}{160}
=0
$