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.
 A: Since $7$ is prime, you know the factors will look like this:$$(7x+a)(x+b)$$
You also know the product of $a$ and $b$ is $-12$, so there are a few possible values.
As a hint, $-25$ is between $7*(-3)$ and $7*(-4)$.
Soon, you will learn about the general solution $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
A: The quadratic formula can help you find roots, from which you can find the equation's factors:
$$x_1, x_2 = \frac{-b \pm \sqrt {b^2-4ac}}{2a} = \frac{25 \pm\sqrt{625-4(7)(-12)}}{2(7)} = \frac{25\pm 31}{14}$$
$x_1 = \dfrac{56}{14} = 4, \quad x_2 = \dfrac{-3}{7}$.
Hence, the equation factors as follows: $$7x^2 -25x - 12 = 0 \iff (x-4)(x+(3/7)) = 0 \iff (x-4)(7x+ 3) = 0$$
A: 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:
$$7(x-4)\left(x+\frac37\right)=0\equiv(x-4)(x+3)=0$$
A: If you have equation $\displaystyle 7x^2 - 25x - 12 = 0$ you can divide both sides by $7$ to get $\displaystyle x^2 - \frac{25x}{7}-\frac{12}{7}=0$. You can do the same in general case $ax^2+bx+c=0$ when $a \neq 0$.
