<s>[INST] What can be the values of $m$ in $(m-3)x^2 + (m+2)x + 2m + 1$? What can be the values of $m$, if $(m-3)x^2 + (m+2)x + 2m + 1$ should be always greater than or equal to zero?
I think we should be using Delta, but I've got no idea how.
 [/INST] Let, 
\begin{align}
y&=ax^2+bx+c\\
&=a\left(x+\dfrac{b}{2a}\right)^2+\left(c-\dfrac{b^2}{4a}\right)\\
&=a\left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{b^2-4ac}{4a}\right)\\
&=a\left[\left(x+\dfrac{b}{2a}\right)^2-\dfrac{D}{4a^2}\right]
\end{align}
From this we can conclude that, when $D\le0$ and $a\ge0$, then only $y\ge0$. Applying this to $y=(m-3)x^2 + (m+2)x + 2m + 1$:
\begin{align}
D&\le0\\
\implies(m+2)^2-4(m-3)(2m+1)&\leq0\\
\implies7m^2-24m-16&\ge0\\
\implies(7m+4)(m-4)&\ge0\\
\implies m\le-\dfrac47\quad\text{and,}\quad m&\ge4\tag{i}
\end{align}
Again, $a\ge0\implies m-3\ge0\implies m\ge3.\tag{ii}$
Therefore from (i) and (ii) we get, $m\ge4$.
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