Question 194110
Set it equal to {{{0}}} to find the factors
{{{x^2 - 12x - 45 = 0}}}
Completing the square:
Take 1/2 of rhe coefficient of the x term, square it,
then add it to both sides:
First add {{{45}}} to both sides
{{{x^2 - 12x = 45}}}
{{{x^2 - 12x + (12/2)^2 = 45 + (12/2)^2}}}
{{{x^2 - 12x + 144/4 = 45 + 144/4}}}
{{{x^2 - 12x + 36 = 45 + 36}}}
Like magic, the left side becomes a perfect square:
{{{(x - 6)^2 = 81}}}
Now take the square root of both sides
{{{x - 6 = 9}}}
and
{{{x - 6 = -9}}}
from these, I get:
{{{x = 15}}}
{{{x = -3}}}
And now I write them as the factors of {{{0}}}
{{{x - 15 = 0}}}
{{{x + 3 = 0}}}
So,
{{{(x - 15)(x + 3) = 0}}}
Cool, or what?