Total sample space of two dice will be equal to $6\times 6=36$

Now for $x^2+\alpha x+\beta >0$ its discriminant $D$ must be less than zero,

So $D=b^2-4ac<0$ or ${\alpha }^2-4\beta <0$$\Rightarrow {\alpha }^2<4\beta$ 

Now, if $\alpha =1,\beta$ can take values $1,2,3,4,5,6$

If $\alpha =2,\beta$ can take values $2,3,4,5,6$

If $\alpha =3,\beta$ can take values $3,4,5,6$

If $\alpha =4,\beta$ can take values $5,6$

If $\alpha =5$ and $6$, no value of $\beta$ possible

So number of favourable ways $=17$

Required probability$=\frac{\text{favourable ways}}{\text{total sample space}}$$=\frac{17}{36}$