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

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

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

We move all terms to the left:

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


Domain of the equation: 5x!=0

x!=0/5

x!=0

x∈R



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

x∈R


We get rid of parentheses

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

We calculate fractions


4x/10x^2+(-5x)/10x^2-5-7=0

We add all the numbers together, and all the variables

4x/10x^2+(-5x)/10x^2-12=0

We multiply all the terms by the denominator


4x+(-5x)-12*10x^2=0

Wy multiply elements

-120x^2+4x+(-5x)=0

We get rid of parentheses

-120x^2+4x-5x=0

We add all the numbers together, and all the variables

-120x^2-1x=0

a = -120; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-120)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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