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What is the set values of p for which p(x^2+x) < 2x^2 + 6x +1 for all real values of x?

Here is my best attempt:

p(x^2+x) < 2x^2 + 6x +1

px^2 + 2p - 2x^2 - 6x -1 < 0

(p-2)x^2 - 6x + 2p -1 < 0

For real roots b^2 - 4ac >= 0

so 6^2 - 4 * (p-2) * (2p-1) >= 0

36 - (8p^2 - 20p + 8) >= 0

-8p^2 + 20p + 28 >= 0

root 1: (-20 + (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0

root 1: -20/-16 + (1296^1/2)/16

root 1: 400/256 + 1296/256

= 1696/256

root 2: (-20 - (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0

root 2: -20/-16 - (1296^1/2)/16

root 2: 400/256 - 1296/256

= -896/256

That's as far as I have got...

the answer in the textbook is p < -1

Can you please help.... Thanks a lot.

Gary.