Complete the square: $x^2+x+1=\left(x+\frac12\right)^2+\frac34$. For what values of $x$ is this positive?
Added: Here’s an explanation of completing the square. Suppose that you have a quadratic $x^2+ax+b$. You want to write this in the form $(x+c)^2+d$ for some constants $c$ and $d$. We know that $(x+c)^2+d=x^2+2cx+(c^2+d)$, and we want this to be identically equal to $x^2+ax+b$. That is, we want $x^2+2cx+(c^2+d)$ and $x^2+ax+b$ to be the same polynomial. Clearly this means that we must have $2c=a$ and $c^2+d=b$. In particular, we must have $c=\frac{a}2$. I could also solve for $d$, but in practice it’s easier to work it out each time than it is to use a formula.
Knowing now that $c=\frac{a}2$, I write $\left(x+\frac{a}2\right)^2$ as a first approximation to $x^2+ax+b$, and then I multiply it out to get $x^2+ax+\frac{a^2}4$. This approximation gives me the right $x^2$ and $x$ terms, but in general it gives me the wrong constant term, because $\frac{a^2}4$ is rarely equal to $b$. Therefore I have to adjust my approximation $\left(x+\frac{a}2\right)^2$. I do so by subtracting $\frac{a^2}4$ and adding $b$:
$\left(x+\frac{a}2\right)^2-\frac{a^2}4+b=x^2+ax+b\;,$
as desired. In the case of the quadratic $x^2+x+1$, $a=1$, so $c=\frac{a}2=\frac12$, and my first approximation was $\left(x+\frac12\right)^2$. This has a constant term of $\frac14$ instead of the desired $1$, so I knew that I had to add another $\frac34$.
The only remaining issue is what to do when the coefficient of $x^2$ isn’t $1$, i.e., when we’re dealing with $ax^2+bx+c$ with $a\ne 1$. The easiest approach is to factor out the $a$ to get
$a\left(x^2+\frac{b}ax+\frac{c}a\right)\;.$
Then complete the square on $x^2+\frac{b}ax+\frac{c}a$ to get
$a\left(\left(x+\frac{b}{2a}\right)^2+\text{some constant}\right)\;.$