Find values for x such that A is not invertible. For $A$ not to be invertible, i.e. $\det(A) = 0$
$A =$ $$
    \begin{bmatrix}
    3 & 1 & 7-x \\
    3 & 2-x & 4 \\
    4 & 2 & 8 \\
    \end{bmatrix}
    $$
After some row operations (step 1: R1 - R2 => R1 then R2 - R3 => R2 and lastly R1+R2 => R1) I got:
$A =$ $$
    \begin{bmatrix}
    -1 & -1 & 7-x \\
    -1 & -x & -4 \\
    4 & 2 & 8 \\
    \end{bmatrix}
    $$
I computed the determinant.
Then checked when it equals $0$ and for values of $x$ I got    $\dfrac  {1}{4} (19\pm \sqrt{305})$.
Could you please confirm with me whether I am on the right track? Thanks.
 A: I obtain $$detA=-4 x^2 + 6 x + 2=0\implies x=\frac34\pm \frac{\sqrt {17}}4$$
To find the values for which $det A=0$, you could simplify the matrix as follow
$$\begin{bmatrix}
    3 & 1 & 7-x \\
    3 & 2-x & 4 \\
    4 & 2 & 8 \\
    \end{bmatrix}
\to\begin{bmatrix}
    3 & 1 & 1-x \\
    3 & 2-x & -2 \\
    2 & 1 & 0 \\
    \end{bmatrix}
$$
A: We have
$$A = 
    \begin{bmatrix}
    3 & 1 & 7-x \\
    3 & 2-x & 4 \\
    4 & 2 & 8 \\
    \end{bmatrix}
    $$
Step 1: R1-R2 -> R1
$$\text{det}A= \left\vert
    \begin{array}{ccc}
    0 & -1+x & 3-x \\
    3 & 2-x & 4 \\
    4 & 2 & 8 \\
    \end{array} \right\vert
    $$
Step 2: R2-R3 -> R2
$$  \text{det}A=  \left\vert\begin{array}{ccc}
    0 & -1+x & 3-x \\
    -1 & -x & -4 \\
    4 & 2 & 8 \\
    \end{array}\right\vert
    $$
Step 3: R3 + $4$ R2 -> R3
$$\text{det}A=
\left\vert    \begin{array}{ccc}
    0 & -1+x & 3-x \\
    -1 & -x & -4 \\
    0 & 2-4x & -8 \\
    \end{array}\right\vert
    $$
Then, using minors of the first column, we obtain
$$ \text{det}A = (-1) (-1)^{2+1} \left\vert \begin{array}{cc} -1+x &3-x\\ 2-4x &-8 \end{array} \right\vert\\
= 8 - 8x - 6 +2x+12x-4x^2\\
= -4x^2+6x +2\\
=-2 (2x^2 -3x -1)$$
Then, $2x^2 - 3x-1 = 2\left(x-\dfrac{3+\sqrt{17}}{4}\right)\left(x-\dfrac{3-\sqrt{17}}{4}\right)$,
if I haven't made a mistake.
