Question 596729
I would not use the quadratic formula unless I really had to, and the instructions say to use factoring.
 
a) {{{x^2+8x+7=0}}} You factored the quadratic polynomial to get {{{(x+1) (x+7)=0}}}.
For the product to be zero, one factor must be zero.
{{{x+1=0}}} gives you one solution: {{{highlight(x=-1)}}}
{{{x+7=0}}} gives you the other solution: {{{highlight(x=-7)}}}
 
b) {{{2y^2 + 12y +10=0}}} You factored the quadratic polynomial to get  {{{2(y^2+6y+5)=0}}}
Factoring a little further, you get {{{2(y+1)(y+5)=0}}}
One of those factors must be zero.
{{{y+1=0}}} gives you one solution: {{{highlight(y=-1)}}}
{{{y+5=0}}} gives you the other solution: {{{highlight(y=-5)}}}
 
c) {{{x^2=81}}} is easy, except for the question of using factoring.
{{{x^2=81}}} means that x is a number that squared equals 81.
You know one such number: {{{9^2=81}}} and {{{highlight(x=9)}}} is one solution.
There is also {{{highlight(x=-9)}}} because {{{(-9)^2=81}}} too.
Maybe you were expected to transform it by subtracting 81 from both sides to get
{{{x^2-81=0}}} and then factor to get {{{(x-9)(x+9)=0}}}.