Completing the squares

Completing the square is most useful for putting a quadratic into vertex form: $a(x - h)^2 + k$.
In this form the vertex sits at $(h, k)$, so it is the fastest way to find a parabola's turning point.

When the coefficient of $x^2$ is $1$, you just complete the square as usual.
For example, $x^2 + 6x + 5$ becomes $(x + 3)^2 - 9 + 5 = (x + 3)^2 - 4$, so its vertex is $(-3, -4)$.

When the leading coefficient is not $1$, factor it out of the $x$-terms first.
Take $2x^2 + 8x + 1$: factor $2$ from the $x$-terms to get $2(x^2 + 4x) + 1$.

Now complete the square inside the parentheses, then multiply the correction back out.
Inside, $x^2 + 4x$ needs $4$, so $x^2 + 4x = (x + 2)^2 - 4$.
Then $2[(x + 2)^2 - 4] + 1 = 2(x + 2)^2 - 8 + 1 = 2(x + 2)^2 - 7$.

Once it is in vertex form, just read off $(h, k)$.
Be careful with the sign: $a(x - h)^2 + k$ means the $x$-coordinate is $h$, so $2(x + 2)^2 - 7$ has its vertex at $(-2, -7)$.

Write $2x^2 - 4x - 3$ in vertex form.
Factor $2$ from the $x$-terms: $2(x^2 - 2x) - 3$, and inside, $x^2 - 2x = (x - 1)^2 - 1$.
So $2[(x - 1)^2 - 1] - 3 = 2(x - 1)^2 - 5$.

Find the vertex of $f(x) = 2x^2 - 8x + 5$.
Factor $2$ from the $x$-terms: $2(x^2 - 4x) + 5$, and inside, $x^2 - 4x = (x - 2)^2 - 4$.
So $f(x) = 2(x - 2)^2 - 3$, and the vertex is at $(2, -3)$.

Find the vertex of $y = x^2 + 10x + 21$.
The leading coefficient is $1$, so complete the square directly: half of $10$ is $5$, giving $(x + 5)^2 - 25 + 21$.
So $y = (x + 5)^2 - 4$, and the vertex is at $(-5, -4)$.