Why do the conditions $x_1+x_2=b$ and $x_1\cdot x_2=ac$ hold for any quadractic equation? Consider the equation $$ax^2+bx+c=0.$$ The factorization of the left hand side is of the form $(x+x_1)(x+x_2)$, where the solutions $x_1$ and $x_2$ must satisfy $$(1)\quad x_1+x_2=-\tfrac bc\quad\mbox{and}\quad(2)\quad x_1\cdot x_2=\tfrac ca.$$
I solve the equation by putting $$x_1=\tfrac12 b+k\quad\mbox{and}\quad x_2=\tfrac12 b-k.$$ This means (1) holds and by substitution into $(2)$, I get
$$(\tfrac12 b+k)(\tfrac12 b-k)=-ac\iff k^2=ac+\tfrac14 b^2.$$ This yields a solution for $k$, and then I can find $x_1$ and $x_2$. 
I have been using this method with success and I like it a lot, since I always forget the quadratic formula. but I have no clue what I am doing. More specifically, why do the conditions (1) and (2) hold for any quadratic equation?
 A: $(x + x_1)(x + x_2) = x^2 + (x_1 + x_2)x + x_1 x_2$. Now compare coefficients.
A: First of all there are lot of mistakes in your notation the quadratic equation should be factorized as:
$$ax^2 + bx + c = a(x-x_1)(x-x_2)$$
And we want to prove that the following holds: 
$$x_1 + x_2 = -\frac ba \tag{1}$$
and 
$$x_1x_2 = \frac ca \tag{2}$$
From the formula for quadratic equations we have:
$$x_{1/2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
So WLOG let's have:
$$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \quad \text {and} \quad \quad x_1 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$$ 
For $(1)$ just add this together:
$$x_1 + x_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} = - \frac{2b}{2a} - \frac ba \tag{1}$$
For $(2)$ just multiply them together:
$$x_1x_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \times \frac{-b - \sqrt{b^2 - 4ac}}{2a}$$
$$x_1x_2 = \frac{b^2 - (\sqrt{b^2 - 4ac})^2}{4a^2}$$
$$x_1x_2 = \frac{b^2 - b^2 - 4ac}{4a^2} = \frac{c}{a} \tag{2}$$
This relations are know as Vieta's Formulas and back at school, they taught us that this is how Vieta find out this relations.
Actually the Vieta's Formulas don't just hold for polynomials of second degree, they hold for polynomials of any degree. Consider the following polynomial:
$$P(x) = a_nx^n + a_{n-1}x^{n-1} +...a_1x + a_0x^0$$
The Vieta stated that the the $(n − k)^{th}$ coefficient $a_{n−k}$ is related to a signed sum of all possible subproducts of roots, taken $k-at-a-time$. Or using mathematical notation written as:
$$\sum_{1\le i_1\le i_2 \le i_3 ... \le i_k \le n} x_{i1}x_{i2}x_{i3}...x_{ik} = (-1)^k \frac{a_{n-k}}{a_n}$$
