2. Solve the equation $\left(x^{2}+3 x-4\right)^{2}+\left(2 x^{2}-7 x+6\right)^{2}=\left(3 x^{3}-4 x+2\right)^{2}$.
2. The sum of the two factors inside the left parenthesis happens to equal the factor inside the left parenthesis, so we can introduce two parameters to simplify.
Let \(a=x^{2}+3 x-4, b=2 x^{2}-7 x+6\), then \(a+b=3 x^{2}-4 x+2\), the original equation becomes \(a^{2}+b^{2}=(a+b)^{2}\), which means \(a b=\)
0. When \(a=0\), i.e., \(x^{2}+3 x-4=0\), we get \(x_{1}=1, x_{2}=-4\); when \(b=0\), i.e., \(2 x^{2}-7 x+6=0\), we get \(x_{3}=2, x_{4}=\frac{3}{2}\).