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.