To solve the equation \(5y^3 - 34y^2 - 7y = 0\), we first factor out the common term \(y\), giving us \(y(5y^2 - 34y - 7) = 0\). This implies that one solution is \(y = 0\). For the quadratic equation \(5y^2 - 34y - 7 = 0\), we can use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the other solutions, where \(a = 5\), \(b = -34\), and \(c = -7\).
We start with the equation
\[
5y^3 - 34y^2 - 7y = 0.
\]
Factoring out the common term \(y\), we have
\[
y(5y^2 - 34y - 7) = 0.
\]
This gives us one solution:
\[
y = 0.
\]
Next, we need to solve the quadratic equation
\[
5y^2 - 34y - 7 = 0.
\]
Using the quadratic formula
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \(a = 5\), \(b = -34\), and \(c = -7\), we first calculate the discriminant:
\[
b^2 - 4ac = (-34)^2 - 4 \cdot 5 \cdot (-7) = 1156 + 140 = 1296.
\]
Taking the square root of the discriminant, we find
\[
\sqrt{1296} = 36.
\]
Now we can find the two solutions:
\[
y_1 = \frac{34 + 36}{10} = \frac{70}{10} = 7,
\]
\[
y_2 = \frac{34 - 36}{10} = \frac{-2}{10} = -0.2.
\]
The complete set of solutions to the original equation is:
\[
y = 0, \quad y = 7, \quad y = -0.2.
\]
The solutions are
\[
\boxed{0, 7, -0.2}.
\]
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