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

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

is $-\frac{3}{4}$!
Let $x_{1}$ and $x_{2}$ be the roots of the equation. We get two linear equations for their calculation; indeed,

$$
x_{1}+x_{2}=\frac{5}{4} \ldots
$$

and

$$
\frac{x_{1}}{x_{2}}=-\frac{3}{4} \cdots
$$

From the latter, $x_{1}=-\frac{3 x_{2}}{4}$; substituting this into 1):

$$
-\frac{3 x_{2}}{4}+\frac{4 x_{2}}{4}=\frac{5}{4} ; \quad \text { hence } \quad x_{2}=5, \quad \text { so } \quad x_{1}=-\frac{15}{4}
$$

However,

$$
\frac{c}{4}=x_{1} x_{2}=-\frac{15}{4} \cdot 5, \quad \text { so } \quad c=-75
$$

The roots of the resulting equation $4 x^{2}-5 x-75=0$ are:

$$
x_{1,2}=\frac{5 \pm \sqrt{25+16 \cdot 75}}{8}=\frac{5 \pm 35}{8} ; \quad x_{1}=\frac{40}{8}=5, \quad x_{2}=-\frac{30}{8}=-\frac{15}{4}
$$

Gábor Steiner (Toldy Ferenc g. V. o. Bp. II.)