Let's determine the real roots of the following equation:

$$
(x+2)^{2}(x+5)(x-1)=10
$$
I. Solution. Our equation can be written as

$$
\left(x^{2}+4 x+4\right)\left(x^{2}+4 x-5\right)-10=0
$$

The left side can be factored as:

$$
\begin{gathered}
\left(x^{2}+4 x+4\right)\left(x^{2}+4 x-6\right)+x^{2}+4 x+4-10= \\
=\left(x^{2}+4 x-6\right)\left(x^{2}+4 x+4+1\right)=\left(x^{2}+4 x-6\right)\left(x^{2}+4 x+5\right)=0
\end{gathered}
$$

The equation

$$
x^{2}+4 x+5=0
$$

has no real roots because its discriminant is $D=16-20<0$.

The equation

$$
x^{2}+4 x-6=0
$$

yields

$$
x_{1}=-2+\sqrt{10}, \quad \text { and } \quad x_{2}=-2-\sqrt{10}
$$

Imre Csiszár (Bp. I., Petőfi g. IV, o. t.)

II. Solution: We can also arrive at the two quadratic equations obtained in the first solution by multiplying the linear factors and using the substitution

$$
x^{2}+4 x=y
$$

and substituting $y$ into the resulting

$$
(y+4)(y-5)=10, \quad \text { or } \quad y^{2}-y-30=0
$$

equation, whose roots are $y_{1}=6, y_{2}=-5$.

László Perneczky (Kaposvár, Táncsics g. IV. o. t.)

III. Solution: With the substitution $x+2=y$, after rearrangement, we arrive at the equation

$$
y^{4}-9 y^{2}-10=0
$$

from which

$$
y^{2}=10, \quad \text { or } \quad y^{2}=-1
$$

The latter equation has no real roots, while the former gives the roots for $x$ as

$$
x_{1}=-2+\sqrt{10}, \quad x_{2}=-2-\sqrt{10}
$$