Suppose you have a quadratic equation: x² + 5x + 6

You are told it has TWO REAL ROOTS and you are to find them.

The factorised quadratic with two roots will look like this:

(x+a)(x+b)

Where a and b are real numbers (any positive or negative number really)

As this is just another form of the original equation we can say they are equal:

(x+a)(x+b) = x² + 5x + 6

If you mulitply out the brackets using the FOIL (First, Outside, Inside, Last) rule you get:

x² + xa + xb + ab = x² + 5x + 6

It's clear the x² terms cancel out, and if we equate the x terms and the number terms, we are left with

xa + xb = 5x        and  ab= 6

x(a+b) = 5x                  a x b = 6

a + b = 5

So now we must use this information to find a and b.

The factors of 6 are:

6 and 1 

3 and 2

Of those factors, the pair which adds to 5 are 3 and 2.

so a = 3 and b = 2

Now we must check the signs of the factorised equation to check when we multiply ot the bracket we get the original equation again.... 

(x+3)(x+2) = x² + 5x + 6 - CORRECT

Now we use this to find the roots, i.e. the x coordinates were y = 0 

so therefore we make our factorised quadratic equal to 0

(x+3)(x+2)=0

If two things multiplied together = 0, then at least one of them must equal 0...

x+3 = 0 ==> x = -3

OR

x+2 = 0 ==> x = -2

We can check the roots are corrects by replacing a and b with the x terms in the original equation and it should equal 0 for both a and b. 

We now have our roots...

x = -3 

and x = -2