Why a quadratic equations always equals zero? On evaluating quadratic equations, It always equals zero:
$$ax^2+bx+c=0$$
Why zero? Is it possible to use other number for another purpose?
 A: Absolutely! But think about what you end up with. Consider the quadratic equation
$x^2 + 2x + 3 = 2 \, . $
If we now subtract 2 from both sides we get $x^2 + 2x + 1 = 0.$ Meaning that these two equations are just two ways of expressing the same thing. So, to save you the trouble of substracting 2 from both sides, you'll be presented with $x^2 + 2x + 1 = 0$ instead of $x^2 + 2x + 3 = 2.$
In fact, you don't even need a number on the right hand side. What about
$2x^2 + 5x - 9 = x^2 + 3x - 10 \, ? $
I could subtract $x^2 + 3x - 10$ from both sides and end up with our friend $x^2 + 2x + 1= 0$. Any equation of the form $px^2 + qx + r = sx^2 + tx + u$ can be simplified - tidied up, if you will - into the form $ax^2 + bx + c = 0.$ When you come across one in the form $ax^2 + bx + c = 0$ it simply means someone has tidied it all up for you in advance. (And it doesn't change the solutions!)
A: $$ax^2+bx+c=0 \implies ax^2+bx=-c$$
$$ax^2+bx+c-d=0 \implies ax^2+bx+c=d$$
We generally want the quadratic to equal zero, however, because the solutions are the roots of the quadratic.  Roots of functions, i.e. the solutions(s) of functions the form $f(x)=0$ are very important.
A: I FINALLY UNDERSTAND WHY THE QUADRATIC EQUATION IS SET TO ZERO!!!!
You "solve" a quadratic equation by figuring out "WHEN Y=0 what does X equal?"(x,0). That's why you set the equation to 0 and not any other number; to find the X-intercept(s) aka (x, 0) point(s).  
HERE'S WHY:
A quadratic equation can be thought of as a function ("f(x)=" is a fancy way of saying "Y=") ; the point being you have an equation where Y = ax^2 + bx + c
***THE KEY IS THE "Y =" PART!!!
When you're asked how many "solutions"  this type of equation has, it's really asking: how many times does the graph (the Parabola in this case) intersect the X-axis? 
You "solve" a quadratic equation by figuring out "WHEN Y=0 what does X equal?"
That's why you set the equation to 0 and not any other number; to find the X-intercept(s) aka (x, 0) point(s).  
The ROOTS of a quadratic equation exactly means the x-intercepts ((x,0) values)
*NOTE:
No solutions means the parabola does not intercept the x-axis (maybe the parabola is above the x-axis or completely below it)
2 solutions (which is the most common for parabolas) means there are TWO places the parabola crosses the x-axis. 
Finding the solutions means finding the (x,0)point  value(s)
Hope this conceptual understanding helps you as much as it did me! :) 
A: it  is general form  ,namely    second order  polynomial equation   and  express  like
$f(x)=0$ where $f(x)=a*x^2+b*x+c$  what if this is equal to some number $D$?
$a*x^2+b*x+c=D$ so we can write it as $a*x^2+b*x+c-D=0$
A: DaleSwanson's answer is nice. I just include this as an answer because its too long for a comment: 
Consider this, if $a_1,a_2 \neq 0$ then $y=a_1x^2+b_1x+c_1$ and $y=a_2x^2+b_2x+c_2$ give parabolas in the $xy$-plane for particular choices of $b_1,b_2,c_1,c_2$. These parabolas intersect if the equation $a_1x^2+b_1x+c_1 = a_2x^2+b_2x+c_2$ has a solution. Bringing all the terms to the r.h.s yields $(a_2-a_1)x^2+(b_2-b_1)x+c_2-c_1=0$. Let $a=a_2-a_1$, $b=b_2-b_1$ and $c=c_2-c_1$ and we obtain the standard $ax^2+bx+c=0$. Assuming $a \neq 0$ amounts to supposing $a_2 \neq a_1$ and the existence of solutions now characterizes the locations (if any) where the parabolas $y=a_1x^2+b_1x+c_1$ and $y=a_2x^2+b_2x+c_2$ intersect. 
More generally, suppose $y=f(x)$ and $y=g(x)$ are graphs of polynomials with $deg(f)=m$ and $deg(g)=n$. $m<n$, therefore, the number of possible intersections will be at most $n$.
A: The value of c is a simple number with no variable.  So you can move any value on the right side over to the left and it will just become part of c.  Example:
$$x^2+x-6=6$$
$$x^2+x-12=0$$
Therefore, we can set the right hand side equal to any number we want.  We usually set it equal to zero because this helps to solve later.  Example:
$$(x+3)(x-2) = 6$$ vs
$$(x-3)(x+4) = 0$$
The second one is easier to solve because we know anything multiplied by 0 is 0.  That means we can solve each part individually.
EDIT:
After we reach the factored form, we know the answer is in the form of something multiplied by something else equals a number.  If that number is not 0 then we must take both parts into account.  On the other hand, if it is 0 then we can simply ask what will make one of those parts zero?  Then it doesn't matter what the other part is.
$$(x-3)(x+4) = 0$$ 
Here, we know that if $(x-3) = 0$ or $(x+4) = 0$ then the whole thing will equal zero, because anything multiplied by 0 is 0.  So, we can just ask what value of $x$ will make $(x-3) = 0$ true?  
Compare this to the version not set to zero:
$$(x+3)(x-2) = 6$$
Now, we can't make this any simpler.  We must figure out what value of $x$ will make that entire thing true from the start.
