(x/2-2)+(2/3)=(2/x-2)

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

(x/2-2)+(2/3)=(2/x-2)

We move all terms to the left:

(x/2-2)+(2/3)-((2/x-2))=0


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

x∈R


We add all the numbers together, and all the variables

(x/2-2)-((2/x-2))+(+2/3)=0

We get rid of parentheses

x/2-((2/x-2))-2+2/3=0

We calculate fractions


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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

5x^2-2*2x=0

Wy multiply elements

5x^2-4x=0

a = 5; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·5·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*5}=\frac{0}{10}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*5}=\frac{8}{10}
=4/5
$