Answer: A. The solution is {-5,3}
Given:
[tex]\begin{gathered} f(x)=x^2+2x-15 \\ f(x)=0 \end{gathered}[/tex]Let us solve for x when f(x)=0
[tex]\begin{gathered} f(x)=x^2+2x-15 \\ 0=x^2+2x-15 \end{gathered}[/tex]Factor the quadratic equation:
[tex]\begin{gathered} 0=x^2+2x-15 \\ 0=(x+5)(x-3) \end{gathered}[/tex]Solve for x:
[tex]\begin{gathered} 0=(x+5)(x-3) \\ ------------ \\ x+5=0 \\ x=-5 \\ ------------ \\ x-3=0 \\ x=3 \end{gathered}[/tex]Therefore, the solution is {-5,3}