How do you solve the quadratic `3x^2-5x-12=0` using any method?

Method 1 - Completing the square

`0 = 12(3x^2-5x-12)`

`color(white)(0) = 36x^2-60x-144`

`color(white)(0) = (6x)^2-2(6x)(5)+25-169`

`color(white)(0) = (6x-5)2-13^2`

`color(white)(0) = ((6x-5)-13)((6x-5)+13)`

`color(white)(0) = (6x-18)(6x+8)`

`color(white)(0) = (6(x-3))(2(3x+4))`

`color(white)(0) = 12(x-3)(3x+4)`

Hence:

`x = 3" "` or `" "x=-4/3`

`color(white)()`
Method 2 - AC method

Given:

`3x^2-5x-12`

Find a pair of factors of `AC=3*12 = 36` which differ by `B=5`

The pair `9, 4` works.

Use this pair to split the middle term, then factor by grouping:

`0 = 3x^2-5x-12`

`color(white)(0) = (3x^2-9x)+(4x-12)`

`color(white)(0) = 3x(x-3)+4(x-3)`

`color(white)(0) = (3x+4)(x-3)`

Hence:

`x = -4/3" "` or `" "x=3`

`color(white)()`
Method 3 - Quadratic formula

The equation:

`3x^2-5x-12 = 0`

is in the form:

`ax^2+bx+c = 0`

with `a=3`, `b=-5`, `c=-12`

The roots are given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (5+-sqrt((-5)^2-4(3)(-12)))/(2(3))`

`color(white)(x) = (5+-sqrt(25+144))/6`

`color(white)(x) = (5+-sqrt(169))/6`

`color(white)(x) = (5+-13)/6`

That is:

`x = (5+13)/6 = 18/6 = 3" "` or `" "x = (5-13)/6 = -8/6 = -4/3`