Find x that satisfies the equality (matrix determinant): This is the exercise.

Find the value of $x\in\mathbb{R}$ that satisfies
$$
\begin{vmatrix}
 x & -1\\
 3  & 1-x
\end{vmatrix} = 
\begin{vmatrix}
 1 & 0 & -3 \\ 
 2 & x & -6 \\ 
 1 & 3 & x-5 
\end{vmatrix}
$$

This is what I've done.
\begin{align*}
\det (A) &= \det (B)\\
\Rightarrow\det(x-x^{2}+3) &= \det(x^{2}-2x)\\
\Rightarrow x-x^{2}+3&=x^{2}-2x\\
\Rightarrow -2x^{2}+3x+3 &=0
\end{align*}
I've also tried to find $x$ with the quadratic formula but the above equation of determinants isn't equal when I plug in the $x$ I found.
What's wrong? What step do I need to follow?

There's nothing wrong – just that $x=\frac{3\pm\sqrt{33}}4$, so simplifying the determinants after substituting $x$ in might take some work. The determinants on both sides should evaluate to $\frac{9\mp\sqrt{33}}8$.
