<s>[INST] How do you solve quadratic inequalities/functions? 
Find the least value of $n$ such that $(n-2)x^2+8x+n+4 > 0$, where $x  \in \mathbb R$  and  $n \in \mathbb N$.

What I did -
As $x$ is real, $D>0$
$64-4(n+4)(n-2)>0$
$(n-4)(n-6)<0$
$n \epsilon (-6,4)$
But as $n \epsilon N$ 
The least value is 3
According to the book's method,
$n-2>0$
$n>2$
And, 
$64-4(n-2)(n+4)<0$
$(n-4)(n+6)>0$
$n>4$
And least value of n comes out to be 5. 
I don't understand why D<0 when x is real? 
Thanks.
 [/INST] the case $n=0$ gives $$x^2-4x-2<0$$ which is not fulfilled for all real $x$
$$n=1$$ gives $$x^2-8x-5<0$$
$$n=2$$ gives $$8x+6>0$$
for $$n>2$$ we get
$$x^2+2\left(\frac{4}{n-2}\right)x+\left(\frac{4}{n-2}\right)^2+\frac{n+4}{n-2}-\left(\frac{4}{n-2}\right)^2>0$$ and we get
$$\left(x+\frac{4}{n-2}\right)^2+\frac{n^2+2n-24}{(n-2)^2}>0$$
finally we get the answer $$n=5$$ then we have
$$\left(x+\frac{4}{3}\right)^2+\frac{11}{9}>0$$
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