If you remember the binomial theorem at all from Module 3 (let me know if you need more explanation here), but we can use a shortcut for the expansion of \((\sqrt{a}+\sqrt{b})^3\).

In general, \((x+y)^3=x^3+3x^2y+3xy^2+y^3\), so

\((\sqrt{a}+\sqrt{b})^3=a\sqrt a +3a\sqrt b+ 3\sqrt a b+b\sqrt b\)
\((\sqrt{a}+\sqrt{b})^3= (a+3b)\sqrt a+(3a+b)\sqrt b\)

This gets us straight to the form that we need. And since this has to be equal to \(22\sqrt 7+26\sqrt 5\), it is most likely that \(a=7, b=5\), so we can go and check that out. After confirming that it is true, we get that our answer is \(7+5=\boxed{12}\). Let me know if you have any other questions!