Solve the following equation:

$$
5 x-7 x^{2}+8 \sqrt{7 x^{2}-5 x+1}=8
$$
Let $7 x^{2}-5 x=y$, then our equation can be written as

$$
-y+8 \sqrt{y+1}=8
$$

that is,

$$
8 \sqrt{y+1}=8+y
$$

Squaring both sides,

$$
64 y+64=64+16 y+y^{2}
$$

which simplifies to

$$
y-48 y=y(y-48)=0
$$

from which we get

$$
y_{1}=0, \quad y_{2}=48
$$

In the first case,

$$
7 x^{2}-5 x=x(7 x-5)=0
$$

from which

$$
x_{1}=0, \quad x_{2}=\frac{5}{7}
$$

In the second case,

$$
7 x^{2}-5 x-48=0
$$

from which

$$
x_{3}=3, \quad x_{4}=-\frac{16}{7}
$$

Substituting the roots obtained into the original equation, we can verify that all four roots satisfy our equation.

Szabados József (Bp., III., Árpád g. II. o. t.)