Question 584022
{{{12x^2-10x-42=0}}} is equivalent to {{{6x^2-5x-21=0}}}
I would say there are 3 ways to solve it, and each person may have one preferred way.
You could:
factor the polynomial, or
use the quadratic formula, or
"complete the square."
FACTORING (not easy in this case):
Factoring is easier when the coefficient of x^2 is 1. It is easiest if the solutions and coefficients are all integers. It will not work if the solutions are not rational numbers.
We are looking for two factors whose product is {{{6x^2-5x-21=0}}}. The products of factors like {{{(6x-21)(x+1)}}} or {{{(2x-7)(3x+3)}}} will give you the terms  {{{6x^2}}} and {{{-21}}}, but you need to get the term {{{-5x}}} right too. 
If you are lucky, you may "see" the right factors immediately.
If not, there is a procedure.
Multiply the coefficient of {{{x^2}}} times the independent (constant) term, in this case, {{{6*(-21)=-126}}}
Look for pairs of factors for that product
-126=-(1)(126)=-(2)(63)=-(3)(42)=-(6)(21)=-(7)(18)=-(9)(14)
Find a pair of factors that adds to the coefficient of x. In this case, -14 and 9 work, because -14+9=-5).
Re-write the polynomial with two terms in x with the coefficients found.
{{{6x^2-5x-21=6x^2-14x+9x-21}}}
Group the 4 terms in pairs with common factors.
{{{6x^2-14x+9x-21=(6x^2+9x)+(-14x-21)}}}
Extract common factors, twice:
{{{(6x^2+9x)+(-14x-21)=3x(2x+3)+(-7)(2x+3)=(3x-7)(2x+3)}}}
So {{{12x^2-10x-42=0}}} is equivalent to {{{6x^2-5x-21=0}}} and  is equivalent to {{{(3x-7)(2x+3)=0}}}
The solutions are what makes one or both factors zero:
{{{3x-7=0}}} --> {{{3x=7}}} --> {{{highlight(x=7/3)}}}
{{{2x+3=0}}} --> {{{2x=-3}}} --> {{{highlight(x=-3/2)}}}
QUADRATIC FORMULA (complicated, but always works)
An equation of the form {{{ax^2+bx+c=0}}} has the solutions {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}.
In this case a=6, b=-5, and c=-21, so
{{{x = (-(-5)+- sqrt((-5)^2-4*6*(-21)))/(2*6)=(5+- sqrt(25+504))/12=(5+- sqrt(529))/12=(5+- 23)/12}}} --> {{{x=(5+23)/12=28/12=highlight(7/3)}}} or
{{{x=(5-23)/12=-18/12=highlight(-3/2)}}}
COMPLETING THE SQUARE
Transform the equation so that only the terms in {{{x}}}} and {{{x^2}}} are on one side of the equal sign, and the term in {{{x^2}}} is just {{{x^2}}}.
{{{6x^2-5x-21=0}}} --> {{{6x^2-5x=21}}} --> {{{x^2-5/6x=21/6}}} --> {{{x^2-5/6x=7/2}}}
Add to both sides the number that will make the left side a perfect square, and then write it as a square, and simplify:
{{{x^2-5/6x+25/144=7/2+25/144}}} --> {{{(x-5/12)^2=7*62/(2*62)+25/144}}} --> {{{(x-5/12)^2=504/144+25/144}}} --> {{{(x-5/12)^2=529/144}}} --> {{{(x-5/12)^2=(23/12)^2}}}
If {{{(x-5/12)^2=(23/12)^2}}}, either {{{x-5/12=23/12}}} or {{{x-5/12=-23/12}}} 
{{{x-5/12=23/12}}} -->{{{x=(5+23)/12=28/12=highlight(7/3)}}}
{{{x-5/12=-23/12}}} -->{{{x=(5-23)/12=-18/12=highlight(-3/2)}}}