6.254. Let $\alpha$ and $\beta$ be the roots of the equation $3 x^{2}+7 x+4=0$. Without solving this equation, form a new quadratic equation with numerical coefficients, whose roots are $\frac{\alpha}{\beta-1}$ and $\frac{\beta}{\alpha-1}$.
Solution.

By Vieta's theorem, we have $\left\{\begin{array}{l}\alpha+\beta=-\frac{7}{3} \\ \alpha \cdot \beta=\frac{4}{3} .\end{array}\right.$

Let $y^{2}+p x+q=0$ be the new quadratic equation. From the condition by Vieta's theorem, we get

$$
\begin{aligned}
& \left\{\begin{array}{l}
p=-\left(\frac{\alpha}{\beta-1}+\frac{\beta}{\alpha-1}\right)=-\left(\frac{\alpha^{2}-\alpha+\beta^{2}-\beta}{(\beta-1)(\alpha-1)}\right)=-\frac{\alpha^{2}+\beta^{2}-(\alpha+\beta)}{\alpha \beta-(\alpha+\beta)+1} \\
q=\frac{\alpha}{\beta-1} \cdot \frac{\beta}{\alpha-1}=\frac{\alpha \beta}{\alpha \beta-(\alpha+\beta)+1}
\end{array} \Leftrightarrow\right. \\
& \left\{\begin{array}{l}
p=\frac{(\alpha+\beta)^{2}-2 \alpha \beta-(\alpha+\beta)}{\alpha \beta-(\alpha+\beta)+1} \\
q=\frac{\alpha \beta}{\alpha \beta-(\alpha+\beta)+1}
\end{array}\right.
\end{aligned}
$$

Since $\alpha+\beta=-\frac{7}{3}$ and $\alpha \beta=\frac{4}{3}$, we get $p=-\frac{23}{21}, q=\frac{2}{7}$.

The desired equation is

$y^{2}-\frac{23}{21} y+\frac{2}{7}=0 \Leftrightarrow 21 y^{2}-23 y+6=0$.

Answer: $21 y^{2}-23 y+6=0$.