<s>[INST] Factoring a quadratic with number in front of $x^2$ I have not yet understood how to factor a quadratic that contains a number in front of $x^2$, without using the quadratic equation. I am used to just brute forcing numbers such that AB and A + B are solutions to it.
Example:
$7x^2 = 25x + 12$
....
$7x^2 - 25x - 12 = 0$
How steps do I need to take to proceed from here ?
Cheers.
 [/INST] Method I: Completing the square:
$$7x^2 - 25x - 12 
=7\left(x^2-\frac{25}7x-\frac{12}7\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{12}7+\frac{25^2}{14^2}\right)\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{961}{196}\right)\right)\\
=7\left(\left(x-\frac{25}{14}\right)^2-\left(\frac{31^2}{14^2}\right)\right)\\
=7\left(x-\frac{25}{14}-\frac{31}{14}\right)\left(x-\frac{25}{14}+\frac{31}{14}\right)\\
=7(x-4)\left(x+\frac{3}{7}\right)\\
=(x-4)(7x+3)\\
$$

Method II: Quadratic formula:
Using quadratic formula, roots are:
$$\frac{25\pm\sqrt{625+336}}{14}=\frac{25\pm31}{14}=4,\frac{-3}7$$
$$7x^2 - 25x - 12 = 0\equiv(x-4)\left(x+\frac37\right)=0$$
Now just multiply coefficient of $x^2$ as:
$$7(x-4)\left(x+\frac37\right)=0\equiv(x-4)(x+3)=0$$
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