56/5x+3=40/4x-3

Solution for 56/5x+3=40/4x-3 equation:

56/5x+3=40/4x-3

We move all terms to the left:

56/5x+3-(40/4x-3)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 4x-3)!=0

x∈R


We get rid of parentheses

56/5x-40/4x+3+3=0

We calculate fractions


224x/20x^2+(-200x)/20x^2+3+3=0

We add all the numbers together, and all the variables

224x/20x^2+(-200x)/20x^2+6=0

We multiply all the terms by the denominator


224x+(-200x)+6*20x^2=0

Wy multiply elements

120x^2+224x+(-200x)=0

We get rid of parentheses

120x^2+224x-200x=0

We add all the numbers together, and all the variables

120x^2+24x=0

a = 120; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·120·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*120}=\frac{-48}{240}
=-1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*120}=\frac{0}{240}
=0
$