Points of intersection of two functions 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
$

$x(-x^{2}+3x+1)=3\quad$ 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 $x=3$.
