Solve $3x^2+13x+4=0$. The root farther from zero is how many times the root nearer to zero?

$x=\frac{-13\pm \sqrt{169-48}}{6}=\frac{-13\pm 11}{6}=-\frac{2}{6}=-\frac{1}{3}$ or $\frac{-24}{6}=-4$. Both roots are negative because b and c are positive, and $-4÷(-\frac{1}{3})=12$. Dividing the other way round gives $\frac{1}{12}$, and reading only the constant term gives 4.