Find the value of $\left | b-c \right |$ Given that $a, b, c \in \mathbb{Z}$, $a>10$ and
$$(x-a)(x-12)+2=(x+b)(x+c)$$
Find the value of $\left | b-c \right |$ 
NOTE: The answer to this problem (as given on the last page of my book) is $1$ and not an expression in terms of $a$, $b$ or $c$.

I compared the coefficients on both sides of the equation and I was getting two equations (but three unknowns).So I used the fact that $a>10$ and tried to set up inequalities hoping to get something 
  like L.H.S>$k$, R.H.S. so that I could assert that both L.H.S. and R.H.S. equal to k+1 (number between them since we are dealing in Integers).But my tries were in vain.
  Any help would be appreciated.
  Thank you for your time and patience.

 A: Expanding the quadratic, we have $x^2-(12+a)x+2=x^2+(b+c)x+bc$.
Setting coefficients equal, we have $-12-a=b+c$ and $12a+2=bc$.
Solving for $a$ in both equations, $a=-12-b-c=\frac{bc-2}{12}$. Multiplying by 12, we have $bc+12b+12c+144=(b+12)(c+12)=2$.
The constraint $a>10$ gives $bc>122$, and $b,c<0$.
If $b$ and $c$ are integers, then we must have $b+12$ and $c+12$ as $\pm1$ and $\pm2$. Therefore $b$ and $c$ are either $-10$ and $-11$, or $-13$ and $-14$. The first option violates the $bc>122$ constraint, but it doesn't matter, because in either case $|b-c|=1$. The constraint $a>10$ is actually unnecessary for this problem.
A: We need the difference between the roots of the quadratic polynomial $(x-b)(x-c)$. 
If the quadratic polynomial is $Ax^2+Bx+C=0$ then the difference between its roots is $\frac{\sqrt{B^2-4AC}}{A}$. This is because its roots are $\frac{-B\pm\sqrt{B^-4AC}}{2A}$.
In our case the left-hand side is the polynomial $x^2-(a+12)x+12a+2$.
So the difference between its roots, which are $b$ and $c$, is $\sqrt{(a+12)^2-4(12a+2)}$.
I will assume the $I$ in your question means integers. 
Notice that the polynomial inside the square root is $a^2-24a+136=(a-12)^2-8$. We need this to be a square $t$. So $8=(a-12)^2-t^2=(a+t-12)(a-t-12)$.
Now, we consider the factorizations of $8$. $\pm1\times\pm8$,$\pm2\times\pm4$. We give these values to $a+t-12$ and $a-t-12$ and solve for $t$ and $a$.
The option $\pm1\times\pm8$ cannot be because we get fractional $a$. So, we consider the second two. We get $t=1$ and $a=(6+24)/2$.
So, $|b-c|=t=1$.
