Starting with the original equation, we move the constant term of the quadratic over to the other side, so we can set up our "completing the square."

$\displaystyle x^2 - 4x - 16 = 5$

$\displaystyle x^2 - 4x - 16 + 16 = 5 + 16$

$\displaystyle x^2 - 4x = 21$

Since the x^2 coefficient is already 1, we don't have to do any division. We can go straight to completing the square by dividing the x coefficient by 2, squaring the result, and adding that result to both sides.

$\displaystyle \frac{-4}{2} = -2$

$\displaystyle (-2)^2 = 4$

$\displaystyle x^2 - 4x + 4 = 21 + 4$

$\displaystyle x^2 - 4x + 4 = 25$

Since the left side is now a perfect square, we can rewrite it as a squared binomial.

$\displaystyle (x-2)^2 = 25$

Now take the square root of both sides and solve for x.

$\displaystyle \sqrt{(x-2)^2} = \sqrt{25}$

$\displaystyle x-2 = \pm 5$

$\displaystyle x = 2 \pm5$

We start by moving the constant term of the quadratic over to the other side of the equation, in order to set up our "completing the square" form.

$\displaystyle y = 4x^2 + 24x + 16$

$\displaystyle y - 16 = 4x^2 + 24x + 16 -16$

$\displaystyle y -16 = 4x^2 + 24x$

Next we divide both sides by 4 so that the x^2 coefficient will be 1. That will allow us to complete the square.

$\displaystyle \frac{y-16}{4} = \frac{4x^2 + 24x}{4}$

$\displaystyle \frac{y}{4} - 4 = x^2 + 6 x$

Now we are ready to complete the square. We divide the x coefficient by 2, square the result, and add that result to both sides.

$\displaystyle \frac{6}{2} = 3$

$\displaystyle 3^2 = 9$

$\displaystyle \frac{y}{4} - 4 + 9 = x^2 + 6 x + 9$

$\displaystyle \frac{y}{4} +5 = x^2 + 6 x + 9$

Because the right side is now a perfect square, we can rewrite it as a squared binomial.

$\displaystyle \frac{y}{4} + 5 = (x+3)^2$

To finish, all we have to do now is solve for y. We'll subtract 5 to both sides, then multiply by 4.

$\displaystyle \frac{y}{4} + 5 -5 = (x+3)^2 - 5$

$\displaystyle \frac{y}{4} = (x+3)^2 - 5$

$\displaystyle 4*\frac{y}{4} = ((x+3)^2 - 5)*4$

$\displaystyle y = 4(x+3)^2 - 5*4$

$\displaystyle y = 4(x+3)^2 - 20$

The first step is to make sure the x² term has a coefficient of 1. Since we have that, we move onto the next step.

Next, move the "loose" number over to the other side.

$\displaystyle x^{2}-8x=-15$

Now divide the coefficient of the x term by 2 (don't forget the sign!). Add the square of this number to both sides.

$\displaystyle -8x \rightarrow -\frac{8}{2}=-4 \rightarrow (-4)^{2}=16$

$\displaystyle x^{2}-8x+16=-15+16$

Simplify:

$\displaystyle x^{2}-8x+16=1$

Now the left side of the equation can be simplified to a squared factor. The factor that will be squared is going to be x plus the original x coefficient divided by 2 as we calculated above.

$\displaystyle (x-4)^{2}=1$

Take the square root of both sides:

$\displaystyle x-4=\pm 1$

Solve for x:

$\displaystyle x-4=1 \rightarrow x=5$

$\displaystyle x-4=-1 \rightarrow x=3$

$\displaystyle x=3,5$

The first step when completing the square is to move the constant to the other side of the equation by subtracting from both sides.  Don't forget the sign!

When completing the square, the lead coefficient should be one.  To achieve this, divide both sides of the equation by the coefficient of the squared term.

$\displaystyle (2x^2-5x)\div2=x^2-\frac{5}{2}x$

and

$\displaystyle 8\div2=4$

This leaves you with the equation $\displaystyle x^2-\frac{5}{2}x=4$

a. Once the variables are on the left, the constant is on the right, and the lead coefficient is 1, you will create a perfect square trinomial on the left side of the equation.  Do this by starting with the coefficient of the x term.

$\displaystyle \frac{5}{2}$

b. Divide this by 2.

$\displaystyle \frac{5}{2}\div2=\frac{5}{4}$

c. Square this term.

$\displaystyle (\frac{5}{4})^2=\frac{25}{16}$

d. Add the result to both sides of the equation.

$\displaystyle x^2-\frac{5}{2}x+\frac{25}{16}=\frac{89}{16}$

e.  The expression on the left side of the equation is now a perfect square trinomial and can be factored to:

$\displaystyle (x-\frac{5}{4})^2=\frac{89}{16}$

a. Take the square root of both sides.  Don't forget to account for both positive and negative answers.

$\displaystyle \sqrt{(x-\frac{5}{4})^2}=\pm\sqrt{\frac{89}{16}}$

b. Simplify both sides

$\displaystyle x-\frac{5}{4}=\pm\frac{\sqrt{89}}{4}$

c. Isolate the variable by adding 5/4 to both sides.

$\displaystyle x=\frac{5}{4}\pm\frac{\sqrt{89}}{4}$

a. Once the variables are on the left side of the equation and the constant is on the right, make the lead coefficient 1 by dividing both sides of the equation by the coefficient of the squared term.

$\displaystyle (3x^2+12x)\div3=-5\div3$

b. Simplify

$\displaystyle x^2+4x=-\frac{5}{3}$

To complete the square, you must create a perfect square trinomial on the left side of the equation. 

a. Take the square root of both sides.  Don't forget to account for both positive and negative answers.

$\displaystyle \sqrt{(x+2)^2}=\pm\sqrt{\frac{7}{3}}$

b. Simplify both sides, if possible.

$\displaystyle x+2=\pm\sqrt{\frac{7}{3}}$

c. Isolate the variable by subtracting 2 from both sides.

$\displaystyle x=-2\pm\sqrt{\frac{7}{3}}$