The following examples come from RegentsPrep.org and MathisFun.com
Practice Problems
(A quadratic equation is a polynomial equation of degree two. There’s no magic to solving quadratic equations. Quadratic equations can be solved by factoringand also by graphing. 
Let’s do a quick review of factoring.
(If you need a more indepth look at factoring, check the Factoring Section of this site.)
There are primarily three types of factoring: 

*Common Monomial 
ab + ac = a(b + c) 
*Difference of Squares  x^{2 }– 9 = (x +3)(x – 3) 
*Quadratic Trinomial  x^{2 }– 5x + 6 = (x – 3)(x – 2) 
If you can factor, you will be able to solve factorable quadratic equations.
Let’s see how it is done.
Solve for x:
Here are the steps you should follow:
1.  Move all terms to the same side of the equal sign, so the equation is set equal to 0.  This places the equation in standard form. 
2.  Factor the algebraic expression.  (x + 3) and (x + 2) are called factors. These are factors of the expression x^{2} – x – 6. 
3.  Set each factor equal to 0. (This process is called the “zero product property”. If the product of two factors equals 0, then either one or both of the factors must be 0.) 

4.  Solve each resulting equation.  x = 3 and x = 2 are called roots. These are roots of the equation x^{2} – x – 6 = 0. 
Example 1 
Solve for x: x^{2 }+ 3x = 0
Factor the common monomial.  x(x + 3)=0  
Set each factor equal to 0 and solve for x. 
x = 0 
x + 3 = 0 x = 3 

List all values of x.  x = {0, 3} 
Example 2 
Solve for y: y^{ 2} = 16
Get all terms on the same side.  y ^{2 }– 16 = 0  
Factor the difference of squares.  (y + 4)(y – 4) =0  
Set each factor equal to 0 and solve for y. 
y + 4 = 0 
y – 4 = 0 y = 4 

List all values of y.  y = {4, 4} 
Example 3 
Solve for c: c^{ 2 }– 12 = c
Get all terms on the same side.  c^{ 2 }– 12 – c =0  
Arrange the terms in standard form.  c ^{2 }– c – 12 = 0  
Factor the quadratic trinomial.  (c + 3)(c – 4) = 0  
Set each factor equal to 0 and solve for c. 
c + 3 = 0 
c – 4 = 0 c = 4 

List all values of c.  c = {3, 4} 
Example 4 
Solve for x:
Employ “product of the means = product of the extremes” (crossmultiply) for this proportion.  
Get all terms on the same side.  x ^{2 }– 1296 = 0  
Factor the difference of squares.  (x + 36)(x – 36) =0  
Set each factor equal to 0 and solve for x. 
x + 36 = 0 
x – 36 = 0 

List all values of x.  x = {36, 36} 
Example 5 
Solve for x:
Employ “product of the means = product of the extremes” (crossmultiply) for this proportion.  
Get all terms on the same side.  0 = 2x ^{2 }– 6x – 8  
Factor the common monomial.  0 = 2(x ^{2 }– 3x – 4)  
Factor the quadratic trinomial.  2(x – 4)(x + 1) =0  
Set each factor equal to 0 and solve for x. 
x + 1 = 0 

List all values of x.  x = {1, 4} 
Example 6 
Write a quadratic equation, in the form ax^{2 }+ bx + c = 0, whose roots are 2 and 5.
The simplest answer will be an equation where the factors of the expression are (x – 2) and (x – 5). Create this equation. 
(x – 2)(x – 5) = 0 
Multiply.  x ^{2 }– 5x – 2x + 10 = 0 
Combine to get an answer equation.  x ^{2 }– 7x + 10 = 0 
Example 7 
The square of a number exceeds 5 times the number by 24. Find the number(s).
Translate the problem into a mathematical equation.  x^{2} = 5x + 24  
Get all terms on the same side.  x ^{2 }– 5x – 24 = 0  
Factor the difference of squares.  (x – 8)(x + 3) =0  
Set each factor equal to 0 and solve for x. 
x – 8 = 0 
x + 3 = 0 

List all values of x.  x = {8, 3} 
Example 8 
In football, the height of the football reached during a pass can be modeled by the equation h = 16t ^{2} + 28t + 6, where the height, h, is in feet and the time, t, is in seconds. How long does it take for this ball to reach a height of 12 feet?
Substitute 12 into the equation for h.  12 = 16t^{ 2} + 28t + 6  
Get all terms on the same side. Move terms to the left side to avoid working with a negative leading coefficient.  16t^{ 2} – 28t + 6 = 0  
Factor the quadratic trinomial.  (4t – 1)(4t – 6) =0  
Set each factor equal to 0 and solve for t. 
4t – 1 = 0 
4t – 6 = 0 

List all values of t that are positive. Negative time, should it appear, is not considered an answer.  t = {1/4, 3/2}. Reaches a height of 12 feet when time is 0.25 seconds (ball going up) and 1.5 seconds (ball coming down). 
Graphing Quadratic Equations
A Quadratic Equation in Standard Form
(a, b, and c can have any value, except that a can’t be 0.)
Here is an example:
Graphing
You can graph a Quadratic Equation using our Function Grapher, but to really understand what is going on, you can make the graph yourself. Read On!
The Simplest Quadratic
The simplest Quadratic Equation is:
f(x) = x^{2}
And its graph is simple too:
This is the curve f(x) = x^{2}
It is a parabola.
Now let us see what happens when we introduce the “a” value:
f(x) = ax^{2}
 Larger values of a squash the curve
 Smaller values of a expand it
 And negative values of a flip it upside down
Play With ItNow is a good time to play with the “Quadratic Equation Explorer” so you can see what different values of a, b and c produce 
The “General” Quadratic
Before graphing we rearrange the equation, from this:
f(x) = ax^{2} + bx + c
To this:
f(x) = a(xh)^{2} + k
Where:
 h = b/2a
 k = f( h )
In other words, calculate h (=b/2a), then find k by calculating the whole equation for x=h
First of all … Why?

So …
 h shows you how far left (or right) the curve has been shifted from x=0
 k shows you how far up (or down) the curve has been shifted from y=0
Lets see an example of how to do this:
Example: Plot f(x) = 2x^{2 }– 12x + 16
First, let’s note down:
 a = 2,
 b = 12, and
 c = 16
Now, what do we know?
 a is positive, so it is an “upwards” graph (“U” shaped)
 a is 2, so it is a little “squashed” compared to the x^{2 }graph
Next, let’s calculate h:
And next we can calculate k (using h=3):
So now we can plot the graph (with real understanding!):
We also know: the vertex is (3,2), and the axis is x=3
From A Graph to The Equation
What if you have a graph, and want to find an equation?
Example: you have just plotted some interesting data, and it looks Quadratic:
Just knowing those two points we can come up with an equation.
Firstly, we know h and k (at the vertex):
(h, k) = (1,1)
So let’s put that into this form of the equation:
f(x) = a(xh)^{2} + k
f(x) = a(x1)^{2} + 1
Then we calculate “a”:
We know (0, 1.5) so:  f(0) = 1.5  
And we know the function (except for a):  f(0) = a(01)^{2} + 1 = 1.5  
Simplify:  f(0) = a + 1 = 1.5  
a = 0.5 
And so here is the resulting Quadratic Equation:
f(x) = 0.5(x1)^{2} + 1
Note: This may not be the correct equation for the data, but it’s a good model and the best we can come up with.