Question 103336
To set this up in base form for quadratic equations it must look like this...
ax^2+bx+c=0
So, your equation becomes: x^2+8x+16=0
In factored form it becomes: (x+4)(x+4)=0        
Both solutions will be the same in this case...but this will usually not be the case.

(x+4)=0
x=-4 or x=-4
I am a little rusty but I think is is called a "double root".

Plug the solution -4 in for every instance of x to prove the solution is valid.
4^2+8(-4)+16=0
16+-32+16=0
-16+16=0
0=0