If the numbers appeared on the two throws of a fair six faced die are $\alpha$ and $\beta$, then the probability that $x^2+\alpha x+\beta >0$, for all $x\in R$, is:

$\begin{array}{l}{x}^2+\alpha x+\beta >0,\mathrm{\forall }x\in R\\ D={\alpha }^2-4\beta <0\\ {\alpha }^2<4\beta \end{array}$

Total cases $=6\times 6=36$

Fav. cases $=\beta =1,\alpha =1$

                $\begin{array}{l}\beta =2,\alpha =1,2\\ \beta =3,\alpha =1,2,3\\ \beta =4,\alpha =1,2,3\\ \beta =5,\alpha =1,2,3,4\\ \beta =6,\alpha =1,2,3,4\end{array}$

                Total favourable cases = 17

$P\left(x\right)=\frac{17}{36}$