complex numbers for quadratic equation Find the quadratic equation whose roots are $2+i$ and $3-i$. 
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
 A: In general, if you want a polynomial $p(x)$ over $\mathbb{C}$ of degree $n$ with roots $a_1,...,a_n$ we have $p(x)=(x-a_1)\cdots(x-a_n)$ (where $a_i$ are not necessarily distinct - corresponding to roots of multiplicity). It is clear that each $a_i$ is in fact a root because $p(a_i)=(a_i-a_1)\cdots(a_i-a_i)\cdots(a_i-a_n)=0$.
A: If $a, b$ are roots of a quadratic equation, then the quadratic equation can be written as
$$(x-a)(x-b) = 0$$
This is intuitively true because substituting in either of the roots $a,b$ will cause LHS to yield $0$.
If you want a more rigorous proof, this result is known as the Weierstrass factorization theorem, which states that any entire function can be represented by a product involving its zeroes.
In your case, this is just 
$$(x-(2+i))(x-(3-i)) = 0$$
You can expand it out if you want it in the form $ax^2 + bx + c = 0$.
A shortcut to the expansion would be to notice that $(x-a)(x-b) = x^2 - (a + b)x + ab$. The coefficient of $x$ is negative of the sum of the roots, while the term independent of $x$ is the product of the roots. This concept is further generalized to higher degree polynomials with Vieta's Formulas.
