Find the nature of the roots of the equation $ 6x^4-25x^3+81x^2-9x-13=0$ using Descartes's rule of signs

$f(x)=(2x-1)(3x+1)(x^2-4x+13)$

so you have roots at $x=\frac{1}{2}, x=\frac{-1}{3}, x=\frac{4\pm\sqrt{16-52}}{2}$

or after tidying

$x=\frac{1}{2}, x=\frac{-1}{3}, x=2\pm3i$