How to solve this determinant equation in a simpler way

Notice that the first two columns are proportional, hence linearly dependent, if

$$\frac x 2 = \frac 4 x$$

which is the same as $x = \pm 2 \sqrt{2}$ after solving. Convince yourself that the third column can never be written as a linear combination of the first two by noticing that the first and second columns have equal components in rows 1 and 3; you can work this into rigorous proof that the above equation gives all solutions.

Subtract the first row from the last and use Laplace's formula for the third row: $$\begin{vmatrix} x & 2 & 3 \\ 4 & x & 1 \\ x & 2 & 5 \\ \end{vmatrix}= \begin{vmatrix} x & 2 & 3 \\ 4 & x & 1 \\ 0 & 0 & 2 \\ \end{vmatrix} = 2(-1)^{3+3}\begin{vmatrix} x & 2\\ 4 & x \end{vmatrix} = 2(x^2-8) = 2(x-2\sqrt{2})(x+2\sqrt{2})=0.$$ Now the roots are $x = \pm 2\sqrt{2}$.

Just expand the determinant! We want $$ 5x^2+2x+24-2x-40-3x^2=0 $$ which simplifies to $2x^2-16=0$.

(In general, finding determinants via row/column relations is faster when a matrix is large. But $3 \times 3$ matrices aren't that large yet...)