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Three main ways to solve quadratic equations.
1) Factoring. Requires some guess and check, you'll get better at it with practice, but you just need to recognize certain patterns and even then it's mostly useless for anything with non-integer solutions
2) Completing the square. This is a simple method that, as far as I know isn't really used beyond a week in algebra because it can be generalized into...
3) The quadratic equation. This is the most simple method in many ways, it will always work even if you extend the quadratic into complex numbers.
Even though the quadratic method is generally superior, factoring does have the advantage that, if you are good at it you can solve quadratics even faster than with the quadratic equation.
Some tips for factoring:
Recognize major patterns like
>x^2 - n^2 = (x + n)(x - n) (Difference of Two Squares: DoTS)
>x^2 + 2nx + n^2 = (x + n)^2
>x^2 - 2nx + n^2 = (x - n)^2
>x^2 + nx = (x + n)(x)
Assume that m and n are two positive integers and m ≤ n
>If b and c are both positive then the factorization will be in the form (x + m)(x + n)
>If b is negative and c is positive then (x - m)(x - n)
>If b and c are both negative then (x + m)(x - n)
>If b is positive and c is negative then (x - m)(x + n)
For the most part factoring is useful when a is 1, if it's not 1 then try dividing all coefficients by a
Factor the absolute value of c into its possible factor pairs. If b and c are both positive or negative then see if there are any factor pairs that add up to the absolute value of b, if one is negative and one is positive then see if any factor pairs have a difference of |b|.
Example:
>3x^2 + 33x - 36
First factor out a to get
>3x^2 + 33x - 36 = 3(x^2 + 11x -12)
12 has factor pairs of
(1, 12)
(2, 6)
(3, 4)
Because b is positive and c is negative, we want to look at the differences which are
12 - 1 = 11
6 - 2 = 4
4 - 3 = 1
Obviously (1, 12) works! b is positive and c is negative so we get
>3x^2 + 33x -36 = 3(x - 1)(x + 12)
