Solutions to the inequality $0>-x^2 +2x+3.$ I am trying to solve an inequality of 
$$0>-x^2 +2x+3.$$
I am aware of two different ways of factorizing this.
$$(-x+3)(x+1)\quad\text{ and }\quad(x-3)(-x-1).$$
When I use $(-x+3)(x+1)$, I get the desired solution of $x>3$ and $x<-1$. However when I use $(x-3)(-x-1)$, this gives me $x<3$ and $x<-1$. How is this possible?

1) $0 > -x^2 + 2x + 3$ so 
$0> (-x+3)(x+1)$
So $(-x + 3)(x+1)$ is negative.  So one of the terms is positive and the other is negative.
So either $-x +3 > 0$ AND $x+1 < 0$
OR
$-x + 3 < 0$ AND $x+1 > 0$
If  $-x +3 > 0$ AND $x+1 < 0$ then $x < 3$ and $x < -1$.  Notice these are redundant statements.  If $x < -1$ then OF COURSE $x < 3$.  There's no need to state $x < 3$.  So 
So if  $-x +3 > 0$ and $x+1 < 0$ then  $x < -1$.
OR
If  $-x +3 < 0$ AND $x+1 > 0$ then $x > 3$ and $x > -1$.  Notice these are redundant statements.  If $x > 3$ then OF COURSE $x > -1$.  There's no need to state $x > - 1$.  So 
So if  $-x +3 < 0$ and $x+1 > 0$ then  $x > 3$.
So the solution is when EITHER $x > 3$ OR $x < -1$.  It can not be when both as you stated as they are contradictory statements.
2) $0 > -x^2 + 2x + 3$ so
$0 > (x-3)(-x-1) $
So EITHER $x-3 < 0$ and $-x-1 > 0$
OR
$x-3 > 0$ and $-x-1 < 0$
So either both $x < 3$ and $x < -1$ so $x < -1$
OR 
both $x > 3$ and $x > -1$ so $x > 3$.
1) and 2) give the exact same results.
