20y² + 27y + 9 = 0

Multiply 20 times 9, getting 180

Factor 180 into primes 2·2·3·3·5

These can be grouped (2·2·5)(3·3) = 20·9,
      but, alas, 20+9 is not 27 L
So try grouping them another way (2·3·5)(2·3) = 30·6,
      but alas, 30+6 is not 27 L
So try grouping them another way (2·2·3)(3·5) = 12·15,
      and hooray, 12+15 really is 27!!! J

So we go back to

 20y² + 27y + 9 = 0

and rewrite the middle term using the numbers 12 and 15
So we rewrite 27y as 12y + 15y

   20y² + 12y + 15y + 9 = 0

Factor 4y out of the first two terms and 3 out of the 
last two:

 4y(5y + 3) + 3(5y + 3) = 0

Now factor (5y + 3) out of each of those

       (5y + 3)(4y + 3) = 0

Set the first factor = 0;   Set the second factor = 0

       5y + 3 = 0               4y + 3 = 0
           5y = -3                  4y = -3 
            y = -3/5                 y = -3/4

So the solutions are -3/5 and -3/4

Edwin