In general, for a equation of the type:

a x \^ 2 + bx + c = 0

there are different ways to solve them depending on whether b and c are 0.

If b = 0, for example x \^ 2 - 9 = 0, you can just rearange it to: x \^ 2 = 9. Then take the sqrt on both sides to get x = +-3.

Now if c = 0, for example x \^ 2 - 2x = 0, you can always factor out the x and get to: x \* ( x - 2) =0. Now this product is 0 if either x = 0, so this is our first root, or if x - 2 = 0, which gives us the solutions x = 0 and x = 2.

Now the last case is if b and c are non-zero. Note that completing the square is just a way to reformulate everything using the first or second binomial formula.

For example, if we start with x\^2 + 6x + 7 = 0, we try to add some a to both sides, such that we can actually use the binomial formula. In this case, we have x\^2 + 6x. The binomial formula is (x+c)\^2 = x\^2 + 2c x + c\^2. For us the 6x corresponds to 2 c x, which implies that c = 3. So if we add and subtract c\^2 we have:

x\^2 + 6x + 3\^2 - 3 \^2 + 7 = 0,

Now the first three terms are in the correct form to use the binomial formula and we get:

(x + 3)\^2 - 9 + 7 = 0

(x + 3)\^2 - 2 = 0

And you know how to solve it from here. Note that the equation you linked is already in that form if you multiply it with -1:

(t - 7)\^2 - 4 = 0

So basically, in the example you have provided, somebody has already done the completing the square part, so there's no need for you to do that again.