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The inequality of p$$^2$$ + 5 < 5p + 14 can be satisfied if:
We have,Β p$$^2$$ + 5 < 5p + 14
=>Β p$$^2$$ - 5p - 9 < 0
=> p< $$\ \frac{\ 5\ +\ \sqrt{\ 61}}{2}$$ or p>Β $$\ \frac{\ 5\ -\ \sqrt{\ 61}}{2}$$Β
=> p<6.4 or p>-1.4
Hence, p β€ 6, p > β1 will satisfy the inequalities
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