362. Solve the equation:

$$
\frac{x^{2}+x+2}{3 x^{2}+5 x-14}=\frac{x^{2}+x+6}{3 x^{2}+5 x-10}
$$
$\triangle$ Let's get rid of the denominators in the equation. It is reasonable to consider the sums $x^{2}+x$ and $3 x^{2}+5 x$ in the left and right parts of the equation as single terms:

$$
\frac{\left(x^{2}+x\right)+2}{\left(3 x^{2}+5 x\right)-14}=\frac{\left(x^{2}+x\right)+6}{\left(3 x^{2}+5 x\right)-10}
$$

Now we get:

$\left(x^{2}+x\right)\left(3 x^{2}+5 x\right)-10\left(x^{2}+x\right)+2\left(3 x^{2}+5 x\right)-20=$

$=\left(x^{2}+x\right)\left(3 x^{2}+5 x\right)-14\left(x^{2}+x\right)+6\left(3 x^{2}+5 x\right)-84$, $4\left(x^{2}+x\right)-4\left(3 x^{2}+5 x\right)+64=0, \quad x^{2}+x-3 x^{2}-5 x+16=0$, $2 x^{2}+4 x-16=0, x^{2}+2 x-8=0$.

The roots of the last equation are $-x_{1}=2, x_{2}=-4$. Let's check the original equation: does any of the two denominators in this equation turn to zero? It turns out they do not. Therefore, these are the roots of the original equation.

Answer: $2,-4$.