\(\dfrac{3+4i}{2+i}=\dfrac{(3+4i)(2-i)}{(2+i)(2-i)^{(*)}}=\dfrac{3\cdot2-3\cdot i+4i\cdot2-4i\cdot i}{2^2-i^2^{(**)}}\\\\=\dfrac{6-3i+8i+4}{4-(-1)}}=\dfrac{10+5i}{4+1}=\dfrac{10+5i}{5}=\dfrac{10}{5}+\dfrac{5i}{5}=\boxed{2+i}\to \boxed{a=2}\\\\used:\\(*)\ (a+b)(a-b)=a^2-b^2\\(**)\ i^2=-1\)
