y(y6+y)-5y(y+36)y-55=-100

First you have to expand your terms.

7y^2 - 5y^3 - 180y^2 - 55 = -100

Collect like terms.

-5y^3 -173y^2 = -45

now factor the y^2

y^2(-5y - 173) = -45

There are two solutions to this equation.

y = 0
and
5y + 173 = 45 which yields
y = -25.6

Dan, what you did at the end is totally invalid

Factors can only be set each equal to zero and solved if their product was equal to zero.
Your product of factors was equal to -45

All you have to do is sub in one of your solutions, y = 0, into the equation to see that it cannot possibly work.

The original posters question was typed in a very strange way.
e.g. what does y6 mean, did he/she mean
y^6 or 6y
I see you read it as 6y, if so, then what was the point of the brackets in the first place?

To solve the equation y(y6 + y) - 5y(y + 36)y - 55 = -100, let's break it down step by step:

Step 1: Simplify the equation.
Start by distributing the terms to get rid of the parentheses.

y * y^6 + y * y - 5y * y - 5y * 36 * y - 55 = -100

This simplifies to:

y^7 + y^2 - 5y^2 - 180y^2 - 55 = -100

Step 2: Combine like terms.
Combine the terms that have the same exponent.

y^7 - 4y^2 - 180y^2 - 55 = -100

Step 3: Simplify further.
Combine the terms with the same variable, even if the exponents are different.

y^7 - 184y^2 - 55 = -100

Step 4: Move all terms to one side of the equation.
By adding 100 to both sides of the equation, we can simplify further.

y^7 - 184y^2 - 55 + 100 = 0

y^7 - 184y^2 + 45 = 0

Step 5: Factor the equation.
Factoring the equation may help us find the values of y that satisfy the equation. However, factoring a 7th-degree polynomial can be quite complex. It's possible that finding an exact solution for the equation may not be practical.

However, if you need to find approximate solutions, you can use numerical methods such as Newton's method or graphing the equation to approximate the solutions.

So, in summary, to solve the equation y(y6 + y) - 5y(y + 36)y - 55 = -100, we simplify and rearrange the equation to make it manageable. Then, we either try to factor it to find exact solutions or use numerical methods to approximate the solutions.