Different ways to factor

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 $$(x-2)(2x+3)$$

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.

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

$$\frac{2(3x-2)\cdot3(2x+7)}{6}$$

$$(3x-2)(2x+7)$$