Solve the following inequality $x^2+x+1\gt 0$ 
Solve the following inequality $x^2+x+1\gt 0$

I understand how to solve inequalities and what the graphs look like. Usually the first step is to set this as in equation and then find the zeros. But for this one when I used the quadratic formula my two answers were: 
$$\dfrac{-1+i\sqrt{3}}{2}$$
$$\dfrac{-1-i\sqrt{3}}{2}$$
I do not know what to do after this, or if I messed up in any way. Or, if there is another way to solve this problem. Any hints help. Please do not solve this problem in any way for me. Thanks!
 A: It has no real zeroes and is continuous. What does this allow you to conclude about the sign of the function?
A: 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 
$$a\left(\left(x+\frac{b}{2a}\right)^2+\text{some constant}\right)\;.$$
A: Why don't you try plotting the graph of the function and guess?

