<s>[INST] Solving $(2y-4)(2y+1) = (2y-2)^2$ I'm getting different answer from answer key.
Solving $$(2y-4)(2y+1) = (2y-2)^2$$
FOIL left side $$4y^2+2y-8y-4 = (2y-2)^2$$
Right side $$4y^2+2y-8y-4 = 4y^2+4 $$
Subtract $4y^2$ from both sides
$$2y-8y-4 = 4 $$
Combine $y$
$$6y-4 = 4$$
add 4 to both sides
$$6y = 8$$
But the answer key has $y=4$
 [/INST] The error is in the "Right Side" step.
You essentially wrote
$$(2y-2)^2 = 4y^2 + 4.$$
That's incorrect.
Remember: $(a-b)^2 = a^2 - 2ab + b^2$. So
$$(2y-2)^2 = 4y^2 - 8y + 4.$$
The third displayed equation should thus be
$$4y^2 +2y - 8y - 4 = 4y^2 -8y + 4.$$
You will find that this leads to $2y = 8$, from which you get $y=4$. 
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