Question 203173
This is all about factoring 
 for the first one we have 2x^2 -4x +1 =0
 You need to find two number that multiply 2 but add to negative 4
-2, -2 

now the question will read like this 
2x^2-2x-2x+1=0
To factor this one we group the first two terms together and the last two terms together. 
(2x^2-2x)(-2x+1)=0
Now we factor out what we can from the barkets
In the first one (2x^2-2x) What is common about the two terms? 2 and x
so we remove them 
2x(x-1) 
now do the same for the second set of barkets (-2x+1)
-1 is common 
-(2x-1)
Now the 2x-1 is common so we rewrite the question as this 
(2x-1)(x-1)=0
Solve for x by splitting the barkets up 
2x-1 = 0   OR    x-1=0
slove by rearranging 
2x-1=0               or x-1=0
2x=1                    x=1
x=1/2

If you do not understand factoring and rearraging please say so in next question.

I will answer the third one and you can try the second your own if you don't understand it please re-post with where you seem to go wrong. ie; factoring.

The next one is a little tricky because the question is equal to something (-7)
We must re-write the question then to make it equal zero in order to factor it. 
x^2+8x+9=-7
now will read this 
x^2+8x+9+7=0  REMEMBER when you move a number or term it changes signs a  postive become a negative and vis versa. 
Collect like terms; so you add 9 and 7 to get 16  
x^2+8x+16=0
Now you need to find two numbers that add to 8 but mult. to 16
4 and 4 and you split the middle term (8x) into these two numbers so the question reads like this now.
x^2+4x+4x+16=0 
(x^2 +4x)(4x+16)=0
x(x+4)4(x+4)=0
(x+4)(x+4)=0
or  (x+4)^2
x must be = to -4
x+4=0
x=-4