Question 1830
completing the square...we want the form {{{(x+a)^2}}} or {{{(x-a)^2}}}. These will give expanded polynomials like...

{{{(x+a)^2 = x^2 + 2ax + a^2}}} --1
{{{(x-a)^2 = x^2 - 2ax + a^2}}} --2

so, your polynomial is {{{x^2 + 6x - 7 = 0}}}. Ignore the c value (-7 in this case), we are interested in the first two terms.

IF, i repeat IF we have {{{x^2 + 6x}}} then what would be the final number, c? THAT is the question you need to answer, so lets look at equation 1 above and ours and compare them...

{{{x^2 + 2ax}}}
{{{x^2 +  6x}}}

as you can see, the 2a=6 therefore a must be 3. Now if a is 3 then the {{{a^2}}} term will be 9. So basically, we NEED to have {{{x^2 + 6x + 9}}}

However we have -7 and not 9, so look at the equation below and convince yourself that it is the same polynomial as your original one...

{{{x^2 + 6x + 9 -16}}}

Effectively all i have done is split the -7 into 2 numbers which still add up to give -7, so there is no change in the polynomial. However, I can now use the 9 to re-write the polynomial, as

{{{(x+3)^2 - 16 = 0}}}
{{{(x+3)^2 = 16}}}
{{{(x+3)}}} = +4 or -4
hence x=-7, x=+1


cheers
Jon.