3. Solve the equation $\left(2 x^{2}-3 x+1\right)^{2}=4 x^{2}-6 x$ $+53$
3. Solution The original equation can be transformed into
$$
\begin{array}{l}
\left(2 x^{2}-3 x+1\right)^{2}-2\left(2 x^{2}-3 x+1\right)-3=0, \\
\left(2 x^{2}-3 x+1-3\right)\left(2 x^{2}-3 x+1+1\right)=0, \\
\left(2 x^{2}-3 x-2\right)\left(2 x^{2}-3 x+2\right)=0 .
\end{array}
$$

Thus, we have $2 x^{2}-3 x-2=0$,
(1)

or $2 x^{2}-3 x+2=0$.
From (1), we get $x_{1}=-\frac{1}{2}, x_{2}=2$; from (2),
$\Delta<0$, no real roots.
$\therefore$ The solutions to the original equation are $\mathrm{x}_{1}=-\frac{1}{2}, \mathrm{x}_{2}=2$.
(Alternatively, the substitution method can be used by setting $2 x^{2}-3 x+1=y$ to solve)