The given function is
[tex]h(r)=-(r+9)^2+36[/tex]To find the zeros, we make h(r) = 0.
[tex]\begin{gathered} -(r+9)^2+36=0 \\ -(r+9)^2=-36 \\ (r+9)^2=36 \\ \sqrt[]{(r+9)^2}=\sqrt[]{36} \\ r+9=\pm36 \\ r_1=36-9=27 \\ r_2=-36-9=-45 \end{gathered}[/tex]