Points of intersection of two functions

Question

Through the following steps I found the x-coordinates of the intersection points of two functions:
$(x)= -x^{2}+3x+1\: and\: g(x)=3/x $
The numbers I found are x=(1, 2, 3)
But on the graph, one of the points has a negative x value, could you guys point me to anything I have missed in my calculations.

$-x^{2}+3x+1=3/x \\x(-x^{2}+3x+1)=3 \\x=3 \\and \\-x^{2}+3x+1=3 \\x^{2}-3x-1=-3 \\x^{2}-3x-1+3=-3+3 \\x^{2}-3x+2=0 \\(x-1)(x-2)=0 \\So \\x=1 \:and\: x=2 $

Answer 1

$x(-x^{2}+3x+1)=3\quad$ doesn't imply $x=3$ or $-x^{2}+3x+1=3$. Instead of that we have $$x(-x^{2}+3x+1)=3\iff -x^3+3x^2+x-3=0\iff-(x+1)(x-1)(x-3)=0$$ Therefore the graphs of the functions mets at $x=-1$, $x=1$ and $x=3$.

Answer 2

The first equation is a cubic, $ -(x+1)(x-1) (x-3) = 0, $

with roots $ x = \pm 1, 3 $ at points of intersection.