Question 1026796
Completing the square.

1.)x^2+6x=7

2.) x^2+8x=9

3.) 9x^2+19x=-8 (solve by completing the square an coefficient of x^2 is 1)

I'm looking for the processes of these. The book examples don't help.
<pre>{{{x^2 + 6x = 7}}}
{{{x^2 + 6x + ((1/2)b)^2 = 7 + ((1/2) * b)^2}}} ------ Adding {{{1/2}}} the value of b, and then squaring the result
{{{x^2 + 6x + ((1/2)6)^2 = 7 + ((1/2) * 6)^2}}} ------ Adding {{{1/2}}} ------ Substituting + 6 for b
{{{x^2 + 6x + 3^2 = 7 + 3^2}}} 
{{{(x + 3)^2 = 7 + 9}}}
{{{(x + 3)^2 = 16}}}
{{{sqrt((x + 3)^2) = " "+- sqrt(16)}}} ------- Taking square root of both sides
{{{x + 3 = " "+- 4}}}
{{{x = " "+- 4 - 3}}}
{{{highlight_green(matrix(1, 7, x, "=", 1, or, x, "=", - 7))}}}

Follow the above steps and you should be able to do number 2
Number 3 can get quite MESSY. Did you really mean: {{{9x^2 + 19x = - 8}}}, or something else? If it's correct, factor
out 9 on both sides to get a coefficient of 1 on {{{x^2}}}, and once again, follow the steps in number 1