Let's determine $a$ such that in the equation

$$
x^{2}+a x+2 a=0
$$

one root is four times the other.
If $x_{1}$ and $x_{2}$ are the roots of the equation, then according to the problem, $x_{2}=4 x_{1}$, thus

$$
x_{1}+x_{2}=5 x_{1}=-a
$$

and

$$
x_{1} x_{2}=4 x_{1}^{2}=2 a
$$

From the first equation,

$$
x_{1}^{2}=\frac{a^{2}}{25}
$$

from the second equation,

$$
x_{1}^{2}=\frac{a}{2}
$$

from which

$$
a_{1}=0, \quad a_{2}=\frac{25}{2}
$$

Therefore, the equation is:

$$
x^{2}+\frac{25}{2} x+25=0
$$

(Bauer Elemér, Budapest.)

Number of solutions: 26.