The easy part is algebraic manipulation:
$x - \left( {\sqrt 5 + \sqrt 6 } \right) = 0 \Rightarrow {x^2} = 5 + 6 + 2\sqrt {30} \Rightarrow {x^2} - 11 = 2\sqrt {30} \Rightarrow p\left( x \right) = {x^4} - 22{x^2} + 1 = 0$. Solutions to this equation are ${x_{1,2,3,4}} = \pm \sqrt 5 \pm \sqrt 6 $ and these are all candidates for the conjugates. However, not all of them are necessarily conjugates of ${\sqrt 5 + \sqrt 6 }$.
The hard part is proving that $p$ is the minimal polynomial of ${\sqrt 5 + \sqrt 6 }$ over $\mathbb{Q}$. Eisenstein criterion doesn't give irreducibility when taking $p\left( {x + a} \right)$.
Even though $p$ has no rational zeroes, that still doesn't mean that it is irreducible over $\mathbb{Q}$.
One method which I can think of would be to calculate each of 7 divisors of $p$ and show that they are not in $\mathbb{Q}\left[ X \right]$. However, that would be impractical in the next problem which has a polynomial of degree 31.
How to see that all the candidates are indeed the conjugates of ${\sqrt 5 + \sqrt 6 }$ ?