Why a quadratic equations always equals zero?

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.

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!)

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 $n$.