The sum of the squares of two numbers is $\frac{45}{4}$, the difference between the smaller and the larger is equal to their product. Determine the numbers.

 untranslated portion:
Két szám négyzetének összege $\frac{45}{4}$, kisebbikük és nagyobbikuk különbsége egyenlő a szorzatukkal. Határozzuk meg a számokat. 

(Translated portion only, keeping the original format and line breaks intact):

The sum of the squares of two numbers is $\frac{45}{4}$, the difference between the smaller and the larger is equal to their product. Determine the numbers.
Let the smaller of the two numbers be denoted by $x$, and the larger by $y$. Then $x < y$, since their difference - in the given order - is negative, but a product of two factors can only be negative if the factors have opposite signs. According to the conditions:

$$
x-y=x y, \quad x^{2}+y^{2}=\frac{45}{4}
$$

From these,

$$
x^{2}+y^{2}-\frac{45}{4}=(x-y)^{2}+2 x y-\frac{45}{4}=(x-y)^{2}+2(x-y)-\frac{45}{4}=0
$$

from which

$$
x-y=x y=-\frac{9}{2}
$$

indeed, the other root of the equation is positive. Furthermore,

$$
\begin{gathered}
(x+y)^{2}=\left(x^{2}+y^{2}\right)+2 x y=\frac{45}{4}-9=\frac{9}{4}, \quad \text { hence } \\
x+y= \pm \frac{3}{2} .
\end{gathered}
$$

Now, by adding and subtracting (2) and (1),

$$
2 x= \pm \frac{3}{2}-\frac{9}{2}, \quad 2 y= \pm \frac{3}{2}+\frac{9}{2}
$$

both values of $x$ are negative, and both values of $y$ are positive, so the problem has two solutions, the numbers being:

$$
-3 \text { and } \frac{3}{2}, \quad \text { or }-\frac{3}{2} \text { and } 3
$$