To find the vertex of the parabola, you have to get it into vertex form:

$\displaystyle y=a[b(x-h)]^{2}+k$

The vertex can then be found at the coordinate $\displaystyle (h,k)$.

To get to vertex form, we have to complete the square.

$\displaystyle y=x^{2}-4x+7$

Move the 7 over to the other side by subtracting 7 from both sides of the equation:

$\displaystyle y-7=x^{2}-4x$

You're going to have to add something to both sides of the equation...

$\displaystyle y-7+\square =x^{2}-4x+\square$

...the question now is what. What number, when put in the box, would create a "perfect square" on the right-hand side of the equation?

Well, a perfect square trinomial is one whose factors are the same, like so:

$\displaystyle (x+a)^{2}=x^{2}+2ax+a^{2}$

In other words, we're looking for $\displaystyle a$.

Well, if $\displaystyle a^{2}$ is what goes in the box, and $\displaystyle x^{2}$ is just $\displaystyle x^{2}$, then $\displaystyle -4x$ must equal $\displaystyle 2ax$. Now we can solve for $\displaystyle a$.

$\displaystyle -4x=2ax$

$\displaystyle -4=2a$

$\displaystyle a=-2$

And since $\displaystyle a^{2}$ goes in the box, we need to add 4 to both sides:

$\displaystyle y-7+4 =x^{2}-4x+4$

Now we can factor the right-hand side very neatly:

$\displaystyle y-7+4 =(x-2)^{2}$

After we clean up a bit...

$\displaystyle y-3 =(x-2)^{2}$

...we get:

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

That gives us a vertex of $\displaystyle (2,3)$.

To complete the square, the equation must be in the form:

$\displaystyle \small \small ax^2+2ab+b^2=c$

$\displaystyle \small x^2-8x+9=0$

$\displaystyle \small a=1$

$\displaystyle \small 2ab=-8$

$\displaystyle \small b=-4$

$\displaystyle \small b^2=16$

$\displaystyle \small x^2-8x+9+7=0+7$

$\displaystyle \small x^2-8x+16=7$

$\displaystyle \small (x-4)^2=7$

$\displaystyle \small \small x-4=\pm\sqrt7$

$\displaystyle \small x=4\pm\sqrt7$

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 + 4x - 3 = 0$

$\displaystyle x^2 + 4x = 3$

Then, divide the middle coefficient by 2:

 $\displaystyle \frac{4}{2}=2$

Square that and add it to both sides:

$\displaystyle 2^2 = 4$

$\displaystyle x^2 + 4x + 4 = 3 + 4$

$\displaystyle x^2 + 4x + 4 = 7$ 

Now, you can factor the quadratic:

 $\displaystyle x^2 + 4x + 4 = (x+2)^2$

 $\displaystyle (x+2)^2=7$

Take the square root of both sides:

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

Finish out the solution:

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

$\displaystyle \sqrt{7}-2 = 0.64575131106459$

$\displaystyle -\sqrt{7}-2=-4.64575131106459$

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 + 8x - 12 = 0$

$\displaystyle x^2 + 8x = 12$ 

Then, divide the middle coefficient by 2:

$\displaystyle \frac{8}{2} = 4$

Square that and add it to both sides:

$\displaystyle 4^2 = 16$

$\displaystyle x^2 + 8x +16 = 12 +16$

$\displaystyle x^2 + 8x + 16 = 28$

Now, you can easily factor the quadratic:

$\displaystyle x^2 + 8x + 16 = (x + 4)^2$ 

$\displaystyle (x+4)^2=28$

Take the square root of both sides:

 $\displaystyle x+ 4 = \pm\sqrt{28}$

Finish out the solution:

$\displaystyle x=\pm\sqrt{28}-4$

$\displaystyle \sqrt{28} - 4 = 1.29150262212918$

$\displaystyle -\sqrt{28} - 4 = -9.29150262212918$

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 - 12x - 16 = 0$

$\displaystyle x^2 - 12x = 16$

Then, divide the middle coefficient by 2:

 $\displaystyle \frac{-12}{2}=-6$

Square that and add it to both sides:

 $\displaystyle (-6)^2 = 36$

$\displaystyle x^2 - 12x +36 = 16 + 36$

$\displaystyle x^2 - 12x + 36 = 52$

Now, you can easily factor the quadratic:

$\displaystyle x^2 - 12x + 36 = (x-6)^2$ 

$\displaystyle (x-6)^2=52$

Take the square root of both sides:

 $\displaystyle x- 6 = \pm\sqrt{52}$

Finish out the solution:

$\displaystyle x=\pm\sqrt{52}+6$

$\displaystyle \sqrt{52} + 6 = 13.21110255092798$

$\displaystyle -\sqrt{52} + 6 = -1.21110255092798$

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 - 20x + 31 = 0$ 

$\displaystyle x^2 - 20x =- 31$

Then, divide the middle coefficient by 2:

$\displaystyle \frac{-20}{2}=-10$ 

Square that and add it to both sides:

 $\displaystyle (-10)^2 = 100$

$\displaystyle x^2 - 20x +100 =- 31 + 100$

$\displaystyle x^2 - 20x +100 = 69$

Now, you can easily factor the quadratic:

 $\displaystyle x^2 - 20x +100 = (x - 10)^2$

$\displaystyle (x-10)^2=69$

Take the square root of both sides:

$\displaystyle x - 10 = \pm\sqrt{69}$

Finish out the solution:

$\displaystyle x=\pm\sqrt{69}+10$

$\displaystyle \sqrt{69} +10 = 18.30662386291807$

$\displaystyle -\sqrt{69} +10 = 1.69337613708192$

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 + 5x - 12 = 0$

$\displaystyle x^2 + 5x = 12$

Then, divide the middle coefficient by 2:

 $\displaystyle \frac{5}{2}= 2.5$

Square that and add it to both sides:

$\displaystyle (2.5)^2 = 6.25$

$\displaystyle x^2 + 5x + 6.25 = 12 + 6.25$

$\displaystyle x^2 + 5x + 6.25 = 18.25$

Now, you can easily factor the quadratic:

$\displaystyle x^2 + 5x + 6.25 = (x+2.5)^2$

$\displaystyle (x+2.5)^2=18.25$

Take the square root of both sides:

$\displaystyle x + 2.5 = \pm \sqrt{18.25}$

Finish out the solution:

$\displaystyle x=\pm\sqrt{18.25}-2.5$

$\displaystyle \sqrt{18.25} - 2.5=1.77200187265877$

$\displaystyle -\sqrt{18.25} - 2.5 = -6.77200187265876$

To solve by completing the square, you should first take the numerical coefficient to the “right side” of the equation:

$\displaystyle x^2 + 7x + 15 = 0$

 $\displaystyle x^2 + 7x = - 15$

Then, divide the middle coefficient by 2:

 $\displaystyle \frac{7}{2} = 3.5$

Square that and add it to both sides:

$\displaystyle (3.5)^2 = 12.25$

$\displaystyle x^2 + 7x + 12.25 = - 15 + 12.25$

$\displaystyle x^2 + 7x + 12.25 = -2.75$

Now, you can easily factor the quadratic:

$\displaystyle x^2 + 7x + 12.25 = (x+3.5)^2$

$\displaystyle (x+3.5)^2=-2.75$

Your next step would be to take the square root of both sides. At this point, however, you know that you cannot solve the problem. When you take the square root of both sides, you will be forced to take the square root of $\displaystyle -2.75$. This is impossible (at least in terms of real numbers), meaning that this problem must have no real solution.

From the original equation, we add 18 to both sides in order to set up our "completing the square." 
$\displaystyle 4 = 2x^2 + 12x -18$

$\displaystyle 4+18 = 2x^2 +12x - 18 + 18$

$\displaystyle 22 = 2x^2 + 12x$

To make completing the square sensible, we divide both sides by 2.

$\displaystyle \frac{22}{2} = \frac{2x^2 + 12x}{2}$

$\displaystyle 11 = x^2 + 6x$

We now divide the x coefficient by 2, square the result, and add that to both sides.

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

$\displaystyle 3^2 = 9$

$\displaystyle 11 + 9 = x^2 + 6x + 9$

$\displaystyle 20 = x^2 + 6x + 9$

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

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

Take the square root of both sides, simplify the radical and solve for x.

$\displaystyle \sqrt{20}=\sqrt{(x+3)^2}$

$\displaystyle \pm2\sqrt{5} = x+ 3$

$\displaystyle -3 \pm 2\sqrt{5} = x$

We start by moving the constant term of the quadratic to the other side of the equation, to set up the "completing the square" format.

$\displaystyle 2x^2 -4x + 1 = 3$

$\displaystyle 2x^2 -4x + 1 - 1 = 3 - 1$

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

Now to make completing the square sensible, we divide boths sides by 2 so that x^2 will not have a coefficient. 

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

$\displaystyle x^2 -2x = 1$

Now we can complete the square by dividing the x coefficient by 2 and squaring the result, then adding that result to both sides.

$\displaystyle \frac{-2}{2} = -1$

$\displaystyle (-1)^2 = 1$

$\displaystyle x^2 - 2x + 1 = 1 + 1$

$\displaystyle x^2 - 2x + 1 = 2$

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

$\displaystyle (x-1)^2 = 2$

Take the square root of both sides, and then solve for x.

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

$\displaystyle x-1 = \pm \sqrt{2}$

$\displaystyle x = 1 \pm \sqrt{2}$