> solve(expr2,b) => [b = -(2*sqrt(30)-15)/2,b = (2*sqrt(30)+15)/2]
>
> Of course, you can evaluate the results numerically if you like.
Specifically, in this context that would be
solve(expr2,b); => [b = -(2*sqrt(30)-15)/2,b = (2*sqrt(30)+15)/2]
%, numer; => [b = 2.022774424948339, b = 12.97722557505166]
% = most recent output expression.
Robert Dodier