Question

how is it possible ?

x=(pi +3)/2
2x=pi +3
2x(pi -3)=(pi +3)(pi -3)
2pi*x -6x =(pi)^2 -9
9 -6x =(pi)^2 -2pi*x
9 -6x +x^2 =(pi)^2 -2pi*x +x^2
(3 -x)^2 =(pi -x)^2
3-x = pi -x
pi =3

Answer 1

Everything looks legal, maybe when taking sq roots there is something that gets twisted.

Answer 2

If you take the negative root, u get the start equation.....

Answer 3

look at the 3rd line
if pi = 3 then both sides of equation = 0

Answer 4

Didn't consider the plus or minus when taking the square root. We get
\[3-x= \pm(\pi-x) \implies 3-x=\pi - x~or~3-x=x-\pi\]
The first equation generates the contradiction from the problem. The second generates the value of x given at the get-go.