Since there are no real roots, the graph of $25y^2−6y+1$ cannot cross the horizontal axis and has to be always negative or always positive. Since it is positive for $y=0$, it is always positive.
Since this is the expression for your original equation, this means that you can always find a solution $x$ which gives this value of $y$ (although you should check the trivial fact that this never gives a 0-denominator).
There is also an approach by "amplification" which is really cool. Also the exact same trick works to prove Hölder's inequality and is generally a very important principle for improving inequalities.
It goes like this:
We start out with $$\langle a-b,a-b\rangle\ge 0$$
for $a,b$ in your inner product space, and $a\not=0$, $b\not=0$. This implies
$$2\langle a,b\rangle\le \langle a, a\rangle + \langle b, b\rangle$$
Now notice that the left hand side is invariant under the scaling $a\mapsto \lambda a$, $b\mapsto \lambda^{-1}b$ for $\lambda>0$. This gives
$$2\langle a,b\rangle \le \lambda^2 \langle a,a\rangle + \lambda^{-2}\langle b, b\rangle$$
Now look at the right hand side as a function of the real variable $\lambda$ and find the optimal value for $\lambda$ using calculus (set the derivative to $0$):
$$\lambda^2=\sqrt{\frac{\langle b,b\rangle}{\langle a,a\rangle}}$$
Plugging this value in, we obtain
$$2\langle a,b\rangle\le \sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}+\sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}$$
i.e.
$$\langle a,b\rangle\le\sqrt{\langle a,a\rangle}\sqrt{\langle b,b\rangle}$$
Notice how we took a trivial observation and "optimized" the expression by exploiting scaling invariance.
Best Answer
Your approach is entirely correct, and seems like the best approach to me as well. I'll write out the details here, if only for myself as I don't have any pen and paper at hand.
Bringing all terms to one side, for each value of $p$ we get a quadratic polynomial in $x$, and we want $$(2-p)x^2+6x+(1-2p)>0,$$ for all $x$. This happens if and only if $2-p>0$ and the discriminant of the quadratic is negative, i.e. $$(-6)^2-4(2-p)(1-2p)<0.$$ The latter is a quadratic in $p$, and expanding the products yields $$-4(2p^2-5p-7)<0.$$ By the quadratic formula we see that this inequality holds if and only if $$p<\frac{5-\sqrt{25+56}}{4}=-1\qquad\text{ or }\qquad p>\frac{5+\sqrt{25+56}}{4}=\frac72.$$ We already found the necessary condition that $2-p>0$, so together this shows that $$p(x^2+2)<2x^2+6x+1,$$ if and only if $p<-1$. It seems that you are correct, and your textbook is wrong.
Just as a sanity check; plugging in $p=0$ yields the inequality $$0<2x^2+6x+1,$$ which should hold for all $x$ according to the solution in your textbook. But this inequality fails for $x=-1$.