Question 182224
{{{x^2 + x - 12}}}
If I set the equation equal to {{{0}}}, I can use
the method of completing the square which says:
Take 1/2 of the coefficient of {{{x}}}, square it, 
and add it to both sides
First, add {{{12}}} to both sides
{{{x^2 + x - 12 = 0}}}
{{{x^2 + x = 12}}}

{{{x^2 + x + (1/2)^2 = 12 + (1/2)^2}}}
{{{x^2 + x + 1/4 = 12 + 1/4}}}
{{{(x + 1/2)^2 = 49/4}}}
Take the square root of both sides
{{{x + 1/2 = 7/2}}}
{{{x = 6/2}}}
(a) {{{x = 3}}}
Also I take the negative square root
{{{x + 1/2 = -(7/2)}}}
{{{x = -(8/2)}}}
(b) {{{x = -4}}}
Subtract {{{3}}} from both sides of (a)
{{{x - 3 = 0}}}
Add {{{4}}} to both sides of (b)
{{{x + 4 = 0}}}
Since these are both {{{0}}}, their products is {{{0}}}
{{{(x - 3)(x + 4) = 0}}}
{{{x^2 + x - 12 = 0}}}
The factors are {{{x-3}}} and {{{x+4}}}