How do you solve $n^2-4n-12=0$ by completing the square?

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How do you solve $n^2-4n-12=0$ by completing the square?

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Answer 1 · 2015-07-05T15:41:40

Move the 12 over, then add half the square of -4 to both sides and simplify.

Explanation:

$n^2-4n-12=0$

First, move the 12 over to the right hand side:
$n^2-4n=12$

Second, add half the square of -4 (the coefficient of $n$ ) to both sides:
$n^2-4n+(1/2 * -4)^2=12+(1/2 * -4)^2$
$n^2-4n+(-2)^2=12+(-2)^2$
$n^2-4n+4=12+4$
$n^2-4n+4=16$

This made the left a perfect square, so change it:
$(n-2)^2$ =16 then simplify

$n-2=+-sqrt(16)$
$n-2=+-4$
$n=2+-4$

The solution set is $[-2, 6]$ .

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How do you solve $n^2-4n-12=0$ by completing the square?

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How do you solve n^2-4n-12=0n^2-4n-12=0 by completing the square?

1 Answer

Explanation:

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How do you solve $n^2-4n-12=0$ by completing the square?