okay I got a problem and it says:

If 3+2i is a solution for x2 (x squared ) +mx+n=0, where m and n are real numbers, what is the value of n?

I would not only like an answer but an explanaion as weell, ihave final coming up and ineed to learn this :/

n = (3+2i)(3-2i) = 13

Please stay with me, it is a long explanation.

If 3+2i is a solution, then 3-2i is also a solution. This is because the radical root theorem (I believe it is called the radical root theorem, I am not sure) states that if a polynomial has a+bi as a root, then a-bi is also a root (root is the name for a solution to a polynomial).

The equation x^2+mx+n only has 2 roots. This is because the fundamental theorem of algebra states if a polynomial is x^n, (x to the nth power, in this case n=2) then the equation has n roots (in this case two).
(The fundamental theorem also states that double roots such as (x-1)^2 count as 2 roots and triple roots such as (x-1)^3 count as 3 roots in case you need to know that for you final)

The constant (or n in this case) is 13. This is because there is a theorem that states the product of the roots is equal to k/a (k is the variable representing the constant or n in this equation, a is the coefficent of the first x or x^2 in this equation; however, because there is no coeficent of x^2 in this equation, you don't have to worry about a). (if n is odd then the sum of the products equals -k/a) If you want to test this theorem, set x = to one of the roots and subtract the root from both sides, then multiply them out.
x=3+2i, x=3-2i
x-(3+2i)=0, x-(3-2i)=0
(x-3-2i)(x-3+2i)=0
x^2-3x+2ix-3x+9-6i-2ix+6i+4 (now simplify)
x^2-3x+2ix-3x+9-6i-2ix+6i+4 = x^2-6x+13
13=n

(I did not learn this until I took precalculus).