The way you stated the problem, there is an infinity of possibilities for the other solution. 

► For instance, the quadratic equation: 
   x² – (6 + 4i)x + (9 + 12i) = 0 
has for discriminant: 
   Δ = (6 + 4i)² – 4(9 + 12i) = 36 – 16 + 48i – 36 – 48i = -16 
which is indeed negative. 
Its solutions will then be: 
   x₁ = [(6 + 4i) + 4i]/2 = 3 + 4i 
   x₂ = [(6 + 4i) – 4i]/2 = 3 
And the other solution here is 3. 

► If you are not convinced, the quadratic equation: 
   x² – (6 + 5i)x + (5 + 15i) = 0 
has for discriminant: 
   Δ = (6 + 5i)² – 4(5 + 15i) = 36 – 25 + 60i – 20 – 60i = -9 
which is indeed negative. 
Its solutions will then be: 
   x₁ = [(6 + 5i) + 3i]/2 = 3 + 4i 
   x₂ = [(6 + 5i) – 3i]/2 = 3 + i 
And the other solution here is 3+i. 

► In fact, every quadratic equation of the form: 
   x² – [6 + (4 + α)i]x + (3 + 4i)(3 + αi) = 0 
where α is any real, has for discriminant: 
   Δ = [6 + (4 + α)i]² – 4(3 + 4i)(3 + αi) 
      = 36 – (4 + α)² + 12(4 + α)i – 36 + 16α – 12(4 + α)i 
      = 16α – (4 + α)² 
      = 16α – 16 – 8α – α² 
      = -16 + 8α – α² 
      = -(α – 4)² 
WILL be negative. 
Their solutions will then be: 
   x₁ = [ [6 + (4 + α)i] – (α – 4)i ]/2 = 3 + 4i 
   x₂ = [ [6 + (4 + α)i] + (α – 4)i ]/2 = 3 + αi 
And the other solution will then be is 3+αi. 

Since α can take any real value, you’ll obtain an infinity of solutions of the form 3+αi. 

► So conclusively: 
If the discriminant of a quadratic is negative AND one of the solutions is 3+4i, the only thing we can say about the other solution is that its real part must be 3.