
This is the third page about factoring. It helps you FACTOR QUADRATIC EQUATIONS
In the first page, I presented the perfect square trinomial and the difference of two squares.
In the second page, I presented a method of factoring fourterm expressions with pairs of factors.
In the third, group of pages, I presented methods of factoring trinomials.
Here, I will present the QUADRATIC FORMULA. You may be interested to know the quadratic formula can be derived fairly easily using the concepts of the perfect square trinomial and the difference of two squares.
How to recognize QUADRATIC EQUATIONS 
It has the following form: ax² + bx + c = 0, where a is not zero. 
Before I show you a method of deriving the quadratic formula, would you like to see a page about finding two numbers if you know their sum and their product? This topic is explored in great detail there, along with a little javascript calculator that finds these numbers, along with a complete explanation of how it works.
Here on this page, I will show you a method of deriving the quadratic formula, which is to change the general form of the quadratic equation into an equivalent form that can be factored, assuming b²  4ac ≥ 0:

























This quadratic formula gives us some interesting insights into the form of the graph of the equation,
y = ax² + bx + c
The graph has the shape of a parabola, which is the trajectory a ball follows when it is thrown. If a is positive, the parabola opens upwards. If a is negative, the parabola opens downwards.
The parabola is symmetric about the line x = b/(2a), which as you can see, is the average of the two roots (values of x for which the graph of the equation passes through the xaxis), if it has any.
There are two of roots whenever b² > 4ac. There is one root (or some people say two identical roots) whenever b² = 4ac, and there are no roots if b² < 4ac.
The vertex of the parabolic graph of the equation is the point (h,k) on the line of symmetry through which the parabola passes. Since it's on the line of symmetry, which is x=b/(2a), we know that h=b/(2a). To find the value of k, simply substitute
h = b/(2a)
in the equation
k = ax² + bx + c
And then solve for k.
k = a(b/(2a))² + b(b/(2a)) + c
k = b²/(4a)  b²/(2a) + c
k = b²/(4a)  2b²/(4a) + 4ac/(4a)
k = (b²  4ac)/(4a)
There's that b²  4ac again! If a is positive (parabola opens up) and it has two roots (b² > 4ac) then the y value of the vertex is in negative territory, so you can see that both arms of the graph of the parabola pass through the x axis. Similarly, if a is negative, and b² > 4ac then the vertex has a positive y coordinate, and the downwardopening parabola's arms pass through the x axis. But if b² < 4ac, then the vertex is on the same side of the x axis as the direction of opening, so the parabola does not pass through the x axis. Graph a few of these possibilities to convince yourself of these facts.
The vertex form of the quadratic equation y=ax²+bx+c is
yk = a(xh)²
where
h = b/(2a)
and
k = (b²  4ac)/(4a)
Here's another (quicker) way to derive the vertex form of the quadratic equation.
Start with the quadratic formula:  y = ax² + bx + c 
Subtract c from both sides:  y  c = ax² + bx 
Add b²/(4a) to both sides:  y + b²/(4a)  c = ax² + bx + b²/(4a) 
Factor "a" out of the right side:  y + b²/(4a)  (4ac)/(4a) = a(x² + (b/a)(x) + b²/(4a²)) 
Factor the right side:  y + b²/(4a)  (4ac)/(4a) = a(x + b/(2a))² 
4a is the common denominator:  y + (b²  4ac)/(4a) = a(x + b/(2a))² 
Now let k=(b²  4ac)/(4a) and h=b/(2a) 
y  k = a(x  h)² 
Here's another interesting property of a parabola: It has a "focus", which is a point, and a "directrix", which is a line, such that every point of the parabola is equidistant from the focus and the directrix.
Using the vertex form of the parabola,
y  k = a(x  h)²
where the point (h,k) is the vertex, we define the focus to be the point (h, k + 1/(4a)) and the directrix to be the line y = k  1/(4a). Let me emphasize this:
focus is (h, k + 1/(4a))
directrix is y = k  1/(4a)
Now let's show that every point that satisfies y  k = a(x  h)² is equidistant from the focus and directrix.
First, let's find the distance from the focus to a point (x,y) on the parabola. Since it's on the parabola, point (x,y) satisfies y  k = a(x  h)². Therefore, this point can be expressed (x, a(x  h)²+k)
The distance from the focus (h, k + 1/(4a)) to that point (x, a(x  h)²+k) is given by
d^{2} = (xh)^{2} + (a(xh)^{2}  1/(4a))^{2}
d^{2} = (xh)^{2} + a^{2}(xh)^{4}  (xh)^{2}/2 + 1/(4a)^{2}
d^{2} = a^{2}(xh)^{4} + (xh)^{2}/2 + 1/(4a)^{2}
d^{2} = (a(xh)^{2} + 1/(4a))^{2}
d = a(xh)^{2} + 1/(4a)
Now, the distance from the directrix y = k  1/(4a) to the point (x, a(x  h)²+k) is the difference in the "y" values, a(xh)^{2} + 1/(4a). This distance from the directrix to the point of the parabola is the same as the distance from the focus to that same point.
The proof, above, of the quadratic formula isn't one that you might just conjure up on first thought. In fact, it seems rather arbitrary that certain expressions were added to both sides of the equation. Your math teacher might say, "well, I'm completing the square", but the easier way to remember and repeat the proof (in case you're ever asked to do that) is to write the result in factored form (the secondtolast line of the proof), and then derive each line of the proof from the one just below it  in other words, work backwards from the result.
I used this same technique to show certain hyperbolic equivalences.
The CauchySchwarz Inequality, which is proved using the quadratic formula
Hyperbolic Functions  sinh(x) and cosh(x)
The webmaster and author of this Math Help site is Graeme McRae.