find two consecutive negative odd integers such that the square of the lesser integer is 40 more than the square of the greater integer

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find two consecutive negative odd integers such that the square of the lesser integer is 40 more than the square of the greater integer

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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9 ,11. sounds weird :P
nop
i mean-9,-11

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how did you get that?
it amounts to 81 and 121
i think se said consecutive
wait im sorry buit i dont understand like how you would do that prblem now..ciykd someone explain that? thanks
ok for for a general formula try -x^2 =(-x+2)^2 + 40
but then dont the x's cancel out?
ok
u start with -x^2=(-x+2)^2+40 multiply the square ull get -x^2=x^2-4x+4
sorry -x^2=x^2-4x+4+40
oh ok thank you guys!!
wait where does the 4x come from?

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