Question 390066
{{{(x-(2-5i))(x - (2+5i)) = ((x-2)+5i)((x - 2)-5i))}}}
 = {{{(x-2)^2 - 25i^2 = x^2 - 4x + 4 + 25 = x^2 - 4x + 29 = 0}}}.
Thus any quadratic equation {{{a(x^2 - 4x + 29) = 0}}}, where a is not equal to 0, will have 2 - 51 as root.  
(Note that if A + Bi is a root of the quadratic equation, then so is its conjugate A - Bi.)