Find the integral value of x

Given : $\left|\begin{array}{ccc}x^{2} & x & 1 \\ 0 & 2 & 1 \\ 3 & 1 & 4\end{array}\right|=28$

$\Rightarrow x^{2}(8-1)-x(0-3)+1(0-6)$

$\Rightarrow 8 x^{2}-x^{2}+3 x-6=28$

$\Rightarrow 7 x^{2}+3 x-6=28$

$\Rightarrow 7 x^{2}+3 x-34=0$

$\Rightarrow(7 x+17)(x-2)=0$

$\Rightarrow x=2$

Integral value of $x$ is 2 . Thus, $x=\frac{-17}{7}$ is not an integer.