[tex]\dfrac{4+3\sqrt 5}{4-3\sqrt 5}\\\\\\=\dfrac{\left(4+3\sqrt 5 \right)\left(4+3\sqrt 5 \right)}{\left(4-3\sqrt 5 \right)\left(4+3\sqrt 5 \right)} \\\\\\=\dfrac{\left(4+3\sqrt 5 \right)^2}{4^2 - \left( 3 \sqrt 5 \right)^2}\\\\\\=\dfrac{4^2 + 2\cdot 4 \cdot 3\sqrt 5 + \left( 3\sqrt 5 \right)^2}{16-9\cdot 5}\\\\\\=\dfrac{16+24\sqrt 5 + 45}{16-45}\\\\\\=\dfrac{61+24\sqrt5}{-29}\\\\\\=-\dfrac{61}{29}-\dfrac{24}{29}\sqrt 5\\\\\\=-\left(\dfrac{61}{29}\right) + \left(- \dfrac{24}{29} \right) \sqrt 5[/tex]
[tex]\text{It is now in a form of}~ a+b\sqrt 5~, \text{where}\\ \\a=-\dfrac{61}{29}~~ \text{and}~~ b =- \dfrac{24}{29}[/tex]
