How do you solve $- 4 n^{2} + 2 n - 5 = 0$ $-4n ^ { 2} + 2n - 5= 0$ ?

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How do you solve $- 4 n^{2} + 2 n - 5 = 0$ $-4n ^ { 2} + 2n - 5= 0$ ?

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Answer 1 · 2018-08-09T00:51:07

$n = \frac{1}{4} \pm \frac{\sqrt{19}}{4} i$ $n = 1/4+-sqrt(19)/4i$

Explanation:

We can solve this by completing the square and using the imaginary unit $i$ $i$ , which satisfies $i^{2} = - 1$ $i^2=-1$ . First multiply by $- 4$ $-4$ so we can do much of the arithmetic using integers...

$0 = - 4 (- 4 n^{2} + 2 n - 5)$ $0 = -4(-4n^2+2n-5)$

$0 = 16 n^{2} - 8 n + 1 + 19$ $color(white)(0) = 16n^2-8n+1+19$

$0 = {(4 n - 1)}^{2} + {(\sqrt{19})}^{2}$ $color(white)(0) = (4n-1)^2+(sqrt(19))^2$

$0 = {(4 n - 1)}^{2} - {(\sqrt{19} i)}^{2}$ $color(white)(0) = (4n-1)^2-(sqrt(19)i)^2$

$0 = ((4 n - 1) - \sqrt{19} i) ((4 n - 1) + \sqrt{19} i)$ $color(white)(0) = ((4n-1)-sqrt(19)i)((4n-1)+sqrt(19)i)$

$0 = (4 n - 1 - \sqrt{19} i) (4 n - 1 + \sqrt{19} i)$ $color(white)(0) = (4n-1-sqrt(19)i)(4n-1+sqrt(19)i)$

Hence:

$4 n = 1 \pm \sqrt{19} i$ $4n = 1+-sqrt(19)i$

So:

$n = \frac{1}{4} \pm \frac{\sqrt{19}}{4} i$ $n = 1/4+-sqrt(19)/4i$

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How do you solve $- 4 n^{2} + 2 n - 5 = 0$ $-4n ^ { 2} + 2n - 5= 0$ ?

1 Answer

Explanation:

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How do you solve -4n ^ { 2} + 2n - 5= 0−4n2+2n−5=0-4n ^ { 2} + 2n - 5= 0?

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How do you solve $- 4 n^{2} + 2 n - 5 = 0$ $-4n ^ { 2} + 2n - 5= 0$ ?