Instructions

For the two given equations I and II----

Question 117

I. $$3p^2+15p=-18$$ II. $$q^2+7q+12=0$$

Solution
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$$3p^2+15p+18 = 0$$
$$p^2+5p+6 = 0$$
$$(p+2)(p+3) = 0$$
$$p = -3, -2$$

$$q^2+7q+12 = 0$$
$$(q+4)(q+3) = 0$$
$$q = -4, -3$$

$$p\geq q$$

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