How $x^2 + 5x + 6 \ge 0$ implies $−3 \ge x \ge −2$? How $x^2 + 5x + 6 \ge 0$ implies $−3 \ge x \ge −2$ ?
I represented $x^2 + 5x + 6 \ge 0$ as $(x+3)(x+2) \ge 0$,which means that either $(x+3) \ge 0$ and $(x+5) \ge 0$ which giving $x \ge -3$ or $x \ge -2$ but how does this gives $−3 \ge x \ge −2$?
ADDED: The above expression could also mean that $(x+3) \le 0$ and $(x+2) \le 0$ which will give $x \le -3$ and $x \le -2$,now  should combine this two to get that result? but in that case we would get both $−3 \le x \le −2$ and $−2 \le x \le −3$ (if I am not wrong) but then the first one should be takes as the answer is taken as the second one makes no sense,am I right?
Source of the problem.
Check the solution given there which causes the confusion.

$x^{2}+5x+6=0$, Solutions are $x=-3,x=-2.$
Since $x^{2}+5x+6=(x+3)(x+2)$, we have three cases:


*

*If $x\geq -2$, then $x+3\geq 1$, $x+2\geq 0$, so $x^{2}+5x+6\geq 0.$

*If $x\leq -3$, then $x+3\leq 0$, $x+2\leq -1$, so $x^{2}+5x+6\geq 0.$

*If $-3<x<-2$, then $x+3>0$, $x+2<0$, so $x^{2}+5x+6<0.$


Remark: instead of "$x^2 + 5x + 6 \ge 0$ implies $−3 \ge x \ge −2$", we should say that "$x^2 + 5x + 6 \le 0$ is equivalent to $−3 \le x \le −2$". The double inequality  $−3 \ge x \ge −2$ is a compact form of  $−3 \ge x$ and  $x \ge −2$, which is not satisfied by any real $x$.

