### Instruction: Find all real values of $x$ that satisfy $x + \frac{45}{x-4} = -10.$

Enter all the solutions, separated by commas.### Response: To solve for $x$, I need to get rid of the fraction on the left side.
I can do that by multiplying both sides by $x-4$, as long as $x \neq 4$.
That gives me $x(x-4) + 45 = -10(x-4).$
Expanding the brackets, I get $x^2 - 4x + 45 = -10x + 40.$
Rearranging the terms, I get $x^2 + 6x + 5 = 0.$
This is a quadratic equation, which I can solve by factoring.
I look for two numbers that multiply to $5$ and add to $6$.
Those numbers are $5$ and $1$.
So I can write $x^2 + 6x + 5 = (x+5)(x+1) = 0.$
The solutions are $x = -5$ and $x = -1.$
I check that neither of them is $4$, so I don't have to worry about dividing by zero.
Therefore, the final answer is $x = -5, -1.$