Proving that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ without the quadratic formula and without calculus

Notice that: $$ x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4} \geq 0+\dfrac{3}{4}= \dfrac{3}{4} \ \ .$$

Clearly, if $x\ge 0$ then $x^2+x+1\ge 1>0$. And if $x<0$, then $x^2+x+1>x^2+2x+1=(x+1)^2\ge 0$

As a different sort of argument:

Note that $$x^3-1=(x-1)(x^2+x+1)$$

Thus any root of $x^2+x+1$ is a cube root of $1$. But the only real $n^{th}$ roots of $1$ are $\pm 1$ so the quadratic can have no real roots. This last point is fairly clear, but just in case: $|x|>1\implies \lim_{n\to \infty} |x|^n=\infty$ and $|x|<1\implies \lim_{n\to \infty} |x|^n=0$

As an alternative way to finish, note that $x^3-1$ has one real root, namely $x=1$, and it is montone increasing so it can not have another.