Let's determine the value of $c$ such that the ratio of the roots of the equation

$$
5 x^{2}-2 x+c=0
$$

is $-\frac{3}{5}$.
The relationships between the roots and coefficients of a quadratic equation are as follows:

$$
\begin{gathered}
x_{1}+x_{2}=\frac{2}{5} \ldots \\
x_{1} x_{2}=\frac{c}{5} \ldots
\end{gathered}
$$

The requirement of the problem is:

$$
\frac{x_{1}}{x_{2}}=-\frac{3}{5} \cdots
$$

$x_{1}$ and $x_{2}$ can be calculated from equations 1) and 3): From 3)

$$
x_{1}=-\frac{3}{5} x_{2}
$$

Substituting this into 1):

$$
-\frac{3}{5} x_{2}+x_{2}=\frac{2}{5}, \quad \text { i.e. } \quad \frac{2 x_{2}}{5}=\frac{2}{5}
$$

From 2)

From here, $x_{2}=1$, so $x_{1}=-\frac{3}{5}$.

$$
c=5 x_{1} x_{2}=5 \cdot 1 \cdot\left(-\frac{3}{5}\right)=-3
$$

Therefore, if $c=-3$, then one of the roots is 1, the other is $-\frac{3}{5}$, and their ratio is $-\frac{3}{5}$.

Geza Freud (Berzsenyi Daniel Grammar School, Budapest, Hungary).