jmc

The equation has quadratic form, so complete the square to solve for x. $$x^4-(4b^2-10b)x^2+25b^2=0$$$$x^4-(4b^2-10b)x^2+(2b^2-5b{)}^2-4b^4+20b^3=0$$$$(x^2-(2b^2-5b){)}^2=4b^4-20b^3$$ In order for the equation to have real solutions, $$16b^4-80b^3\ge 0$$$$b^3(b-5)\ge 0$$$$b\le 0\text{ or }b\ge 5$$ Note that $2b^2-5b=b(2b-5)$ is greater than or equal to $0$ when $b\le 0$ or $b\ge 5$. Also, if $b=0$, then expression leads to $x^4=0$ and only has one unique solution, so discard $b=0$ as a solution. The rest of the values leads to $b^2$ equalling some positive value, so these values will lead to two distinct real solutions. Therefore, in interval notation, $b\in [-17,0)\cup [5,17]$, so the probability that the equation has at least two distinct real solutions when $b$ is randomly picked from interval $[-17,17]$ is $\frac{29}{34}$. This means that $m+n=\boxed{63}$.