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

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How do you solve $n^{2} - 4 n - 12 = 0$ $n^2-4n-12=0$ 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:

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

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

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

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

$n - 2 = \pm \sqrt{16}$ $n-2=+-sqrt(16)$
$n - 2 = \pm 4$ $n-2=+-4$
$n = 2 \pm 4$ $n=2+-4$

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

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

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