Verified answer

[tex]\displaystyle\bf\\\frac{3a^{2}+6a }{a^{2} -9}-\frac{2a}{a-3} =\frac{3a^{2}+6a }{(a-3)(a+3)} -\frac{2a}{a-3}=\frac{3a^{2}+6a-2a\cdot(a+3) }{(a-3)(a+3)} =\\\\\\=\frac{3a^{2}+6a-2a^{2} -6a }{(a-3)(a+3)} =\frac{a^{2} }{a^{2} -9} \\\\a=6\\\\\frac{a^{2} }{a^{2} -9} =\frac{6^{2} }{6^{2} -9} =\frac{36}{36-9} =\frac{36}{27}=\frac{4}{3} =1\frac{1}{3} \\\\Otvet: \ 1\frac{1}{3}[/tex]