From Förberedande kurs i matematik 1

We use the standard method and augment the fraction with the conjugate of the denominator \displaystyle \sqrt{5}+2. Then the conjugate rule gives

\displaystyle \begin{align} & \frac{\sqrt{2}+3}{\sqrt{5}-2}=\frac{\sqrt{2}+3}{\sqrt{5}-2}\centerdot \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\left( \sqrt{2}+3 \right)\left( \sqrt{5}+2 \right)}{\left( \sqrt{5} \right)^{2}-2^{2}} \\ & =\frac{\sqrt{2}\centerdot \sqrt{5}+\sqrt{2}\centerdot 2+3\centerdot \sqrt{5}+3\centerdot 2}{5-4}=\sqrt{2\centerdot 5}+2\sqrt{2}+3\sqrt{5}+6 \\ & =6+2\sqrt{2}+3\sqrt{5}+10 \\ \end{align}