Proving that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$ without the quadratic formula and without calculus I'm asked to prove that $\{x\in\Bbb{R}\mid 1+x+x^2 = 0\} = \varnothing$  in an algebra textbook.
The formula for the real roots of a second degree polynomial is not introduced yet. And the book is written without assuming any prior calculus knowledge so I can't  prove this by finding the minimum and the limits as x approaches $\infty \text{ and} -\infty  $. 
So there has to be a simple algebraic proof involving neither the quadratic formula nor calculus but I'm stuck. 
Here are some things I thought:
Method 1:

$1+x+x^2 = 0 \iff 1+x+x^2+x = x$
$\iff x^2+2x+1 = x$
$\iff (x+1)^2 = x $ 
And here maybe prove that there is no x such that $(x+1)^2 = x$ ???

Method 2:

$1+x+x^2 = 0$
$\iff x^2+1 = -x$
By the trichotomy law only one of these propositions hold: $x=0$ or
  $x>0$ or $x<0$.
Assuming $x=0$:
$x^2+1= 0^2+1 = 0 +1 = 1$
$-x = - 0 = 0$
And $1\neq 0$
Assuming $x>0$:
$x>0 \implies -x < 0$
And $x^2+1 \ge 1 \text{  } \forall x$
With this method I have trouble proving the case $x<0$:
I thought maybe something like this could help but I'm not sure:
$x<0 \implies -x=|x|$ 
$x^2 = |x|^2$
And then prove that there is no x such that $|x|^2 + 1 = |x|$??

Can anyone please help me? Remember: No calculus or quadratic formula allowed.

$x=0$ is not a root, so divide by $x \ne 0$ and write the equation as:
$$
x+\frac{1}{x} = -1
$$
This requires $x$ to be negative for the LHS to be negative, but then $y=-x$ is positive and $\displaystyle y+\frac{1}{y} \ge 2$ by AM-GM, so $\displaystyle x+\frac{1}{x} \le -2 \lt -1\,$, therefore there are no real solutions.
