Different ways to factor I'm interested to know some different methods to factor equations of the form $ax^2+bx+c$, where $a \ne 0$, other than pure guess and check. 
 A: Assuming your question is in earnest and you aren't trolling the answer is
$ax^2 + bx + c = a(x  + \frac{b + \sqrt{b^2 - 4ac}}{2a})(x  + \frac{b - \sqrt{b^2 - 4ac}}{2a} )$
No guessing.  If the equation factors at all (which it won't if $4ac > b^2$) then it will always factor to that.
A: Variation of the method mentioned by @suomnonA
1) Multiply $a*c$.
2) Find a factor pair of $ac$ that adds up to $b$. Let them be $x_1$ and $x_2$.
3) Write factors as:
$$\frac{(ax-x_1)(ax-x_2)}{a}$$
The factors on top will factorize down to cancel out with the $a$ on the bottom.
Example:
$$6x^2+17x-14$$
$$a=6, b=17, c=-14$$
Here we do step $1$, getting $$ac=-84$$ Then we do step $2$. The factor pair is $$-4,21$$ Now we do step $3$. 
The equation becomes $$\frac{(6x-4)(6x+21)}{2}$$
which factorizes to
$$\frac{2(3x-2)\cdot3(2x+7)}{6}$$
$$(3x-2)(2x+7)$$
A: Another option:
$$a\,{{x}^{2}}+bx+c$$
$$\downarrow$$
$$n\,{{x}^{2}}-2knx+{{k}^{2}}n-m$$
$$\downarrow$$
$${{x}^{2}}-2kx-\frac{m}{n}+{{k}^{2}}$$
$$\downarrow$$
$$\left( x-\sqrt{\frac{m}{n}}-k\right) \,\left( x+\sqrt{\frac{m}{n}}-k\right) $$
A: One method is The British Method. Here are the steps.
$1)$ Multiply $a*c$.
$2)$ Find a factor pair of $ac$ that adds up to $b$
$3)$ Replace $bx$ in the equation with the two factors, both multiplied by x.
$4)$ Factor by grouping.
Here is an annotated example.
$$2x^2-x-6$$
$$a=2, b=-1, c=-6$$
Here we do step $1$, getting $$ac=-12$$ Then we do step $2$. The factor pair is $$-4,3$$ Now we do step $3$. The equation becomes $$2x^2-4x+3x-6$$
Then we factor by grouping
$$2x^2-4x+3x-6$$ $$2x(x-2)+3(x-2)$$ $$(2x+3)(x-2)$$

Another method is shown below. 
Steps:
$1)$ Multiply $a*c$.
$2)$ Find a factor pair of $ac$ that adds up to $b$, call them $x_1$ and $x_2$
$3)$ Write the following equation: $(ax+x_1)(ax+x_2)$.
$4)$ Factor out the GCF of each individual equation. The part remaining after factoring out the GCF is your answer. 
Annotated example: 
$$2x^2-x-6$$ $$a=2, b=-1, c=-6$$ Here we do step $1$, getting $$ac=-12$$ Then we do step $2$. The factor pair is $$-4,3$$ Now we do step $3$, getting the equation: $$(2x-4)(2x+3)$$ Factor out a $2$ from the first part, where you get the final answer of $$(x-2)(2x+3)$$
