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Solution
$$y^2 + 3y - 18 \geq 0$$
$$\Rightarrow y^2 + 6y - 3y - 18 \geq 0$$
$$\Rightarrow y (y+6) - 3 (y+6) \geq 0$$
$$\Rightarrow (y-3) (y+6) \geq 0$$
$$\Rightarrow y\geq3 and y\leq-6$$
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