Why is this method right to solve some equations of degree 2? Given this equation
$-x^2+x+6=4$
We can write it as
$(3-x)(x+2) = 4$
So weridly if I take
$(3-x) = 4$ and $(x+2) = 4$, I can get $x=-1$ and $x=2$ that are indeed the correct solutions.
Why does this happen? And in which cases can this occur?

Consider $(a-x)(x-b) = c$
With roots $r_1$ and $r_2$.
We need four equalities to be true
$$a-r_1 = c, \quad r_1-b = 1, \quad a-r_2 = 1, \quad r_2-b = c$$
We get
$$\text{$r_1=a-c=b+1$ and $r_2=a-1 = b+c$}$$
Which implies $c=a-b-1$
So for any $a$ and $b$
$$(a-x)(x-b) = a-b-1$$
will behave as you described.
