For the two given equations I and II----
I. $$6p^{2}+5p+1=0$$
II. $$20q^{2}+9q=-1$$
$$6p^2+5p+1 = 0$$
$$(2p+1)(3p+1) = 0$$
$$p = -\frac{1}{2}, -\frac{1}{3}$$
$$20q^2+9q+1 = 0$$
$$(4q+1)(5q+1) = 0$$
$$q = -\frac{1}{4}, -\frac{1}{5}$$
$$p < q$$
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