Can you solve by factoring the quadratic

5y³ + 34y² = 7y 

and show every step?

     5y³ + 34y² = 7y

Get 0 on the right by adding -7y to both sides

5y³ + 34y² - 7y = 0

y is a common factor so factor out y

y(5y² + 34y - 7) = 0

Write this:

y( y    )( y    ) = 0

Now think of a way to write 5, the first
coefficient as the product of two positive 
integers.  Since 5 is prime there is only 
one way, namely 1×5 or 5×1.  So put a 1 
before the first y and a 5 before the second y.  
(It would be OK to reverse these).

y(1y    )(5y    ) = 0

Now think of a way to write 7, the last
term, as the product of two positive 
integers.  Since 7 is also prime there is 
only one way, namely 1×7 or 7×1.  So try 
putting a 1 at the end of the first
parentheses and a 7 at the end of the 
second parentheses: 

y(1y   1)(5y   7) = 0

Now check the OUTERS and INNERS as though
you were going to multiply that out by FOIL,
ignoring the signs.  The OUTER product would
be 1y times 7 or 7y and the INNER product 
would be 1 times 5y or 5y.

But there is no way that 7y and 5y could 
combine to give the mddle term +34y no matter
what signs we gave them.

So let's discard this

y(1y   1)(5y   7) = 0

and reverse the 1 and the 7.

y(1y   7)(5y   1) = 0 

Now again check the OUTERS and INNERS as though
you were going to multiply that out by FOIL,
ignoring the signs.  The OUTER product would
be 1y times 1 or 1y and the INNER product 
would be 7 times 5y or 35y.

This time there IS A way that 1y and 35y could 
combine to give the middle term +34y.  That would
be if we gave the 35y a positive sign and the 1y a
negative sign, since +35y-1y does give the middle
term +34y. So we put a + by the 7 to make the INNERS
be +35y and we put a - by the 1 in the second 
parentheses to make the OUTERS be -1y

y(1y + 7)(5y - 1) = 0

Of course we can erase the 1 before the y in 
the first parentheses:

y(y + 7)(5y - 1) = 0

So the left side is now completely factored.
The factors are y, y + 7, and 5y - 1

We set each factor = 0

Setting the factor y = 0 just gives y = 0

Setting the factor y + 7 = 0, gives y = -7

Setting 5y - 1 = 0 gives 5y = 1 or y = 1/5 

So there are three solutions: 0, -7, and 1/5

Edwin