Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1)  \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I can see that $p^2 > -3(p+3)$ for  almost all values of $p \ $, therefore
$p^2-3(p+3) > 0 $. However, the question asks to show that $\ p^2-3(p+3) \geq 0$  
If I make $p^2 = 3(p+3)$ I can find roots and so $\ p^2-3(p+3) = 0$, when $\displaystyle{p = \frac{3 \pm \sqrt{45}}{2}}$. Therefore  $\ p^2-3(p+3) \geq 0$  
Having done this, how can I mathematically show that $p^2$ is never $<$ than $3(p+3)$? Because I am not satisfied with just saying that by inspection $p^2$ is greater than $3(p+3)$.
Thank you

$$3x^2-2px+(p+3) = 0 \\
9x^2-6px+3(p+3) = 0 \\
(3x-p)^2+(3(p+3)-p^2)=0 \\
p^2-3(p+3) = (3x-p)^2 \ge 0$$

Furthermore you can see that $p^2-3(p+3) = 0$ when $p=3x$ and $x^2 = x+1$, i.e. $x = \frac{1 \pm \sqrt 5}2$
