So the first part of the question is about completing the square. Since the coefficient of .. is greater than one, the first step in carrying out this process is to take out the ‘2’ from the equation. This would look a bit like this: 2[x2 + (9/2)x ] + 1. We’ve left 1 out of this manipulation of the equation just for simplicity, but what has happened to the other parts is that we’ve just divided both by 2. Concentrating on just the x2 + (9/2)x, this is now an easier equation to complete the square with. Start with putting (x + 9/4)2, and taking away (9/4)2. You will get (x + 9/4)2 - (9/4)2, though don’t forget that this is just the middle bit of the equation! Substituting this for x2 + (9/2)x will give you 2[(x + 9/4)2 - (9/4)2 ] + 1, and the next steps just involve multiplying out the brackets: 2(x + 9/4)2 - 81/8 + 1, and then: 2(x + 9/4)2 - 73/8.
This is now written in the form a(x+m)2 + n, and you can get round to solving. The goal is to get the x on its own. Set the equation equal to zero, and start with adding 73/8 to each side. Then, because you can’t do anything with what’s inside the bracket until you deal with the outside, divide by two. You’ll now have: (x + 9/4)2 = 73/16. To get rid of the brackets, then square root both sides- and then just take away 9/4. The answer you will get is x = (√73 - 9)/4, (-√73 - 9)/4