Solve the following equation:

$$
\sqrt{x^{2}+x+2}=x^{2}+x
$$
Let $x^{2}+x=y$. Thus we get

$$
\begin{array}{ll} 
& \sqrt{y+2}=y \ldots \\
\text { or } & y^{2}-y-2=0 \ldots
\end{array}
$$

The roots of 2) are: -1 and 2. However, according to 1), $y>0$ and thus $y=2$. Therefore,

$$
x^{2}+x=2, \quad \text { and from this } \quad x_{1}=1, \quad x_{2}=-2
$$

Indeed: $\sqrt{1+1+2}=1+1$ and $\sqrt{4-2+2}=4-2$.

Huhn László (Kegyesrendi g. VI. o. Szeged)

Note. In an irrational equation, the square root represents only one value, i.e., the positive one.

$y=-1$ satisfies the equation $-\sqrt{y+2}=y$. When $y=-1$,

$$
x^{2}+x=-1, \quad \text { or } \quad x^{2}+x+1=0
$$

The roots of this are complex numbers. Then,

$$
-\sqrt{x^{2}+x+2}=x^{2}+x, \quad \text { that is } \quad-\sqrt{\left(x^{2}+x+1\right)+1}=x^{2}+x
$$

so

$$
-\sqrt{1}=-1
$$