Determine all positive integer $a$ such that the equation $2x^2 - 30x + a = 0$ has two prime roots, i.e. both roots are prime numbers.
1. Let \( p_1 \) and \( p_2 \) (with \( p_1 \leq p_2 \)) be the prime roots of the quadratic equation \( 2x^2 - 30x + a = 0 \). By Vieta's formulas, we know:
   \[
   p_1 + p_2 = \frac{30}{2} = 15
   \]
   and
   \[
   p_1 p_2 = \frac{a}{2}
   \]

2. Since \( p_1 \) and \( p_2 \) are prime numbers and their sum is 15, we need to find two prime numbers that add up to 15. The only pair of prime numbers that satisfy this condition are \( p_1 = 2 \) and \( p_2 = 13 \) (since 2 is the only even prime number and 13 is the only prime number that, when added to 2, gives 15).

3. Substituting \( p_1 = 2 \) and \( p_2 = 13 \) into the product equation:
   \[
   p_1 p_2 = 2 \times 13 = 26
   \]
   Therefore,
   \[
   \frac{a}{2} = 26 \implies a = 52
   \]

4. We verify that \( a = 52 \) satisfies the original equation:
   \[
   2x^2 - 30x + 52 = 0
   \]
   The roots of this equation are indeed \( x = 2 \) and \( x = 13 \), both of which are prime numbers.

Thus, the only positive integer \( a \) such that the equation \( 2x^2 - 30x + a = 0 \) has two prime roots is \( a = 52 \).

The final answer is \( \boxed{52} \).