A simpler way to solve this determinant equation-$2$ 
Question Statement:-
  Solve the following determinant equation 
  $$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}=0$$


My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}=
\begin{vmatrix}
x+13 & x+14 & x+15 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}\tag{$R_1\rightarrow R_1+R_2+R_3$}$$
$$(x+13)\begin{vmatrix}
1 & 1 & 1 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}+
\begin{vmatrix}
0 & 1 & 2 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}\tag{1}$$
Now lets solve the two matrices obtained in the last step above seperately.
$$(x+13)\begin{vmatrix}
1 & 1 & 1 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}$$
$$=(x+13)\begin{vmatrix}
0 & 0 & 1 \\
2 & x & 4 \\
-(x+1) & -x & x+8 \\
\end{vmatrix}\hspace{3cm}(C_1\rightarrow C_1-C_3, C_2\rightarrow C_2-C_3)$$
$$=x(x+13)\begin{vmatrix}
0 & 0 & 1 \\
2 & 1 & 4 \\
-(x+1) & -1 & x+8 \\
\end{vmatrix}=x(x+13)(x-1)$$
And the second matrix can be simplified as 
$$\begin{vmatrix}
0 & 1 & 0 \\
6 & x+4 & -4-2x \\
7 & 8 & x-8 \\
\end{vmatrix}\tag{$C_3\rightarrow C_3-C_2$}$$
$$=-1\times(6x-48+28+14x)=-20(x-1)$$
So, we get 
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}=0\implies x(x+13)(x-1)-20(x-1)=0\\
\implies (x-1)(x^2+13x-20)=0\implies x=1,\dfrac{-13\pm\sqrt{249}}{2}$$
As can be seen, my solution is way long and it is no better than opening the determinant. So if anyone can provide me with a better, simpler and a more intuitive solution I will be very thankful. 

Your computation is correct. 
Here is a different procedure where we use the rule of Sarrus: the determinant is
$$x(x+4)(x+8)+56+144-21(x+4)-32x-12(x+8)\\
=x^3+12x^2-33x+20=(x-1)\cdot(x^2+13x-20).$$
Here you have to guess that $1$ is a root of the third degree polynomial and then divide it by $(x-1)$ in order to obtain the other factor $x^2+13x-20$.
