Solve x^2+6x-7 by completing the square.

We know that to complete the square we put the equation in the form (ax+b)^2+c=0. So we must do that to our equation.

We will begin by establishing what will be within our squared brackets. As there is no coefficient of x^2, we know that a will be 1. However, what is b?

A fast way to find b is to take the coefficient of x and divide it by 2. Here we will be given 3. Consequently, we will provisionally assume that (ax+b)^2 can be written as (x+3)^2.

Now we should test this. To know if we are right, when we expand our brackets the equation should be in the form x^2+6x+c (with c being an integer).

Our (x+3)^2 expands to x^2+6x+9. Now we know our first part is correct. However we are still not in the form where (ax+b)^2+c=x^2+6x-7, as, instead of -7, we are left with +9. As a result we need to take difference.

We take 16 from 9 to get -7. So know we know that (x+3)^2-16=x^2+6x-7. We can now use this to solve the equation as we know that it is equal to 0.

We must remember that when solving equations, you must do the same to both sides of the equation. So, as follows. (x+3)^2-16=0, add 16 to both sides to get: (x+3)^2=16. Now, root both sides to get: x+3=+/-4. We should now take 3 from both sides to get: x=-3+/-4. This should give us two answers: x=-7 and x=1.