Answer
hyperbola
Work Step by Step
The general equation of a conic in the form of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
(i) defines a parabola if $B^2-4AC=0$
(ii) defines an ellipse if $B^2-4AC\lt0$ and $A\ne C$
(iii) defines a circle if $B^2-4AC\lt0$ and $A= C$
(iv) defines a hyperbola if $B^2-4AC\gt0$
Here, we have
$A=2,B=5,C=3$
Hence,
$B^2-4AC=5^2-4(2)(3)=25-24=1\gt0$
Thus, it is a hyperbola.