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Solve for $x$ in the equation $x^{2}+5 x+6 \geq 0$

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$x \leq-3, x \geq-2$

Explanation

(1) Factor $x^{2}+5 x+6$
$(x+2)(x+3) \geq 0$
(2) Solve for $x$.
$x=-2,-3$
(3) From the values of $x$ above, we have these 3 intervals to test.
\begin{aligned} &x \leq-3 \\ &-3 \leq x \leq-2 \\ &x \geq-2 \end{aligned}

(4) Pick a test point for each interval.
For the interval $x \leq-3$
Let's pick $x=-4$. Then, $(-4)^{2}+5 \times-4+6 \geq 0$.
After simplifying, we get $2 \geq 0$, which is true.
Keep this interval..
For the interval $-3 \leq x \leq-2$ :
Let's pick $x=-\frac{5}{2}$. Then, $\left(-\frac{5}{2}\right)^{2}+5 \times \frac{-5}{2}+6 \geq 0$.
After simplifying, we get $-0.25 \geq 0$, which is false.
Drop this interval..
For the interval $x \geq-2$ :
Let's pick $x=0$. Then, $0^{2}+5 \times 0+6 \geq 0$.
After simplifying, we get $6 \geq 0$, which is true.
Keep this interval..
(5) Therefore,
$x \leq-3, x \geq-2$

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