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
