2/5x-7=4/4x+1

Solution for 2/5x-7=4/4x+1 equation:

2/5x-7=4/4x+1

We move all terms to the left:

2/5x-7-(4/4x+1)=0


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



Domain of the equation: 4x+1)!=0

x∈R


We get rid of parentheses

2/5x-4/4x-1-7=0

We calculate fractions


8x/20x^2+(-20x)/20x^2-1-7=0

We add all the numbers together, and all the variables

8x/20x^2+(-20x)/20x^2-8=0

We multiply all the terms by the denominator


8x+(-20x)-8*20x^2=0

Wy multiply elements

-160x^2+8x+(-20x)=0

We get rid of parentheses

-160x^2+8x-20x=0

We add all the numbers together, and all the variables

-160x^2-12x=0

a = -160; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·(-160)·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*-160}=\frac{0}{-320}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*-160}=\frac{24}{-320}
=-3/40
$