Given equation is $2x^3+10x-13=0$.

Given that the eccentricity of a conic satisfies the equation i.e., the eccentricity is any one of the roots.

Now, $2x^3+10x-13=0$.

$\Rightarrow 2x^3+10x=13$

$\Rightarrow x\left(x^2+5\right)=\frac{13}{2}$

$\Rightarrow x\left(x^2+5\right)=6.5$

By verification:

Put $x=0\Rightarrow 0(0+5)=6.5$

Put $x=1\Rightarrow 1\left(1^2+5\right)=6.5$

$\Rightarrow 6=6.5$

Put $x=2\Rightarrow 2\left(2^2+5\right)=6.5$

$\Rightarrow 2\left(9\right)=6.5$

$\Rightarrow 18=6.5$

$\therefore$The value of $x$ lies in between $1, 2$ i.e., the

eccentricity is greater than one.

$\therefore$The conic is hyperbola.