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The inequality of p$$^2$$ + 5 < 5p + 14 can be satisfied if:
p ≤ 6, p > −1
p = 6, p = −2
p ≤ 6, p ≤ 1
p ≥ 6, p = 1
We have, p$$^2$$ + 5 < 5p + 14
=> p$$^2$$ - 5p - 9 < 0
=> p< $$\ \frac{\ 5\ +\ \sqrt{\ 61}}{2}$$ or p> $$\ \frac{\ 5\ -\ \sqrt{\ 61}}{2}$$
=> p<6.4 or p>-1.4
Hence, p ≤ 6, p > −1 will satisfy the inequalities
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