y=x^2 + 6x + 5

what do you need to add to x^2+6x to make it a perfect square?

Well if (x+a)^2=x^2+6x +c for some c then 2a=6, and c=a^2, that is

a=3 and c=9.

So:

y = (x+3)^2 - 9 +5 = (x+3)^2 - 4

(multiply this out to check that it is the same thing that you started with.

So now if you want to solve:

y=x^2 + 6x + 5 = 0

we write:

(x+3)^2 - 4 = 0

or:

(x+3)^2 = 4

so taking square roots:

x+3 = +/- 2

x = -5 or -1.

Which we can check by substituting back into: x^2 + 6x + 5

with x=-5 we have: 25 - 30 +5 =0

and with x=-1 we have: 1 -6 +5 =0

RonL