I have chosen to write an answer attempting to provide the solution for the general case, when "m+√n".
Consider the general order two polynomial with integer coefficients a, b and c.
This is ax^2 + bx + c. We know that this order two polynomial must have two complex roots. One root "m+√n" has been given, and thus the other root may not have an imaginary part. (Otherwise upon expansion, our order two polynomial will most certainly not have integer coefficients). Thus we may re-write our order two polynomials entirely in terms of its roots. This is shown below:
ax^2 + bx + c = a(x-r1)(x-r2). a may not be equal to zero, or else our polynomial will not be order two, so we may divide both sides of our equality by zero. Producing:
(1) x^2 + (b/a)x + (c/a) = (x-r1)(x-r2). We may now deduce that (b/a) = -1(r1+r2) . (b/a) is a member of the rationals, so we must ensure r1+r2 is a member of the rationals. We know r1 = m+√n, and so r1+r2 = m+√n+r2. If r2 is in the rationals, r1+r2 is irrational. Contradiction. r2 must belong to the irrationals. However, we may restrict the value of r2 even further. r2 may ONLY be an expression in the form k - √n where k is in the rationals. This is because we wish to discard the √n term in our r1+r2 expression so as to obtain a rational number. Good, but we still do not know what our value of k should be.
On looking at (1), it is easy to see that r1r2 = c/a. The product of our roots must produce a rational number. We also know that r1 = m+√n and r2 = k-√n . On multiplying r1 by r2, we get, mk + √n(k-m) - n. m,k,n are ALL rationals. Also we know that r1r2 must be rational. Thus we must not have a √n term in our expression (we assume n is non-square, otherwise (m+√n) is simply an integer in which case the problem becomes trivial.) and so k-m = 0. Thus k=m. So we now know that in fact, our second root must be r2 = m-√n . This gives us the quadratic a(x-(m+√n))(x-(m-√n)). Thus r2 is simply the conjugate of r1 which is as expected.