Question 706870
The rational expression {{{(2a^2+a-15)/(5 a^2+16a+3)}}} does not exist when
{{{5 a^2+16a+3=0}}} which happens for {{{a=-1/5}}} and for {{{a=-3}}}
(You can find that by factoring {{{5 a^2+16a+3=(5a+1)(a+3)}}} , or by solving {{{5 a^2+16a+3=0}}} using the quadratic formula).
Those are the only restrictions on the variable {{{a}}}.
For any value of {{{a}}} other than {{{-1/5}}} and {{{-3}}} the expression is defined.