How do you do maths questions like this?

Hint: "multiply by the conjugate"; Multiply top & bottom by (3 + √5), then simplify top&bottom.

You're not guaranteed to get integers, in general, but you can always get rational numbers. Maybe this one will work out to be integers, though...

2(3+√ 5) ÷ (3− √ 5) .... starting point
= [2*(3 + √5)*(3 + √5)] ÷ [(3 - √5)*(3 + √5) ... multiply top and bottom by the conjugate of the bottom
= [2*(9 + 6√5 + 5)] ÷ (9 - 5) .... multiply out top and bottom
= (28 + 12√5) ÷ (4) .... combine "like" terms
= 7 + 3√5 .... divide each term on top by 4

2(3+√5) / (3−√5)
multiply by (3+√5) / (3+√5)
2 (3+√5)² / (3−√5)(3+√5)
2 (9+5+6√5) / (9–5)
2 (14+6√5) / (4)
7 + 3√5

2(3 + √ 5) ÷ (3 − √ 5)
= 7 + 3 sqrt(5)
= 13.70820393249936908922752100619382870632185507883457717281...

You need to remove the 'root sign from the denominator(bottom).
To do this , apply the conjugate, which is ( 3 + sqrt(5))
Hence
2(3 + sqrt(5)(3 + sqrt(5)) / ( 3 - srt(5)(3 + sqrt(5)) +>
Apply FOIL to the brackets
2(9 + 3sqrt(5) + 3sqrt(5) + 5) / ( 9 + 3sqrt(5) - 3sqrt(5) - 5)
Collect like terms
2(14 + 6sqrt(5)) / (4)
(28 + 12sqrt(5)) / 4
7 + 3sqrt(5)

b = 7
c = 3

Done!!!!

You can always "multiply by 1" and it does not change the value.

Of course, you get to pick how to write "1"
Any fraction of "something over itself" is equal to 1 (avoid 0/0 which is undefined).
Here, the use of "conjugates" is helpful.

A conjugate is an expression where the sign of the "weird part" is flipped.
The conjugate of (3 - √5) is (3 + √5)
Use:
1= (3+√5)/(3+√5)
---
2(3+√ 5) / (3− √ 5)
becomes
2(3+√5)/(3−√5) * 1
becomes
2(3+√5)/(3−√5) * (3+√5)/(3+√5)
becomes
2(3+√5)^2 / (3−√5)(3+√5)
becomes
2(3+√5)^2 / (9−5)
becomes
2(3+√5)^2 / 4
becomes
(1/2)(3+√5)^2

It is important to understand that this is the SAME value as the original expression.

We expand the square
(1/2) (9 + 6√5 + 5)
(1/2)(14 + 6√5)
(7 + 3√5)