Since, \( \Large 2+i\sqrt{3} \) is a root of equation \( \Large x^{2}+px+q=0 \) therefore, \( \Large 2-i\sqrt{3} \) will be other root. Now sum of the roots \( \Large 2+i\sqrt{3} \) + \( \Large 2-i\sqrt{3} \) = -P

=> 4 = -P

Product of roots \( \Large (2+i\sqrt{3}) \) \( \Large (2-i\sqrt{3}) \) = q,

=> 7 = -q.

Hence (p,q) = (-4, 7)