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?
 A: 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.
A: Hint: Just be more careful. Write it out in full. Both give you
$$-(x-3)(x+1)<0$$
which is equivalent to
$$(x-3)(x+1)>0$$
This happens when both factors on the LHS have the same nonzero sign.
Can you take it from here?
A: I think you are beginner- so as teacher always says for such errors before doing any calculation just make the highest power of polynomial inequation positive. although the better way to solve it is using number line.  put both factors=0 then on number line mark those point and then for every interval check the validity of equation
