405. Solve the equation:

$$
\sqrt{4 x^{2}+5 x-1}-2 \sqrt{x^{2}-3 x+3}=\frac{17 x-13}{7}
$$
$\triangle$ Let's introduce three new variables:

$$
\sqrt{4 x^{2}+5 x-1}=y, \quad 2 \sqrt{x^{2}-3 x+3}=z, \quad \frac{17 x-13}{7}=t
$$

Then

$$
y-z=t, \quad y^{2}-z^{2}=17 x-13=7 t
$$

It seems natural to divide the second of the obtained equations by the first. But first, let's consider the case $t=0$:

$$
t=0, \quad \frac{17 x-13}{7}=0, \quad x=\frac{13}{17}
$$

Verification shows that the value $x=\frac{13}{17}$ is a root of the original equation.

Let $t \neq 0$. Then

$$
y+z=7, \quad y-z=t
$$

From here, $2 y=7+t$. Express both sides of this equation in terms of $x$ and solve the resulting equation:

$$
\begin{aligned}
& 2 \sqrt{4 x^{2}+5 x-1}=7+\frac{17 x-13}{7} \\
& 14 \sqrt{4 x^{2}+5 x-1}=17 x+36, \quad 196\left(4 x^{2}+5 x-1\right)=(17 x+36)^{2} \\
& 784 x^{2}+980 x-196=289 x^{2}+1224 x+1296 \\
& 495 x^{2}-244 x-1492=0
\end{aligned}
$$

$$
\text { The roots of the last equation are }-x_{1}=2 \text { and } x_{2}=-\frac{746}{495} \text {. }
$$

Let's do the verification. It shows that both roots satisfy the original equation.

Answer: $13 / 17, 2, -746 / 495$.