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#