How do you solve $14 n^{2} + 24 n - 5004 = 0$ $14n^2 + 24n - 5004 = 0$ ?

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How do you solve $14 n^{2} + 24 n - 5004 = 0$ $14n^2 + 24n - 5004 = 0$ ?

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Answer 1 · 2018-03-05T21:47:58

Quadratic formula

Explanation:

The straightforward way is to use the quadratic formula:
$n = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $n=(-b+-sqrt(b^2-4ac))/(2a)$
This is for expressions of the form $a n^{2} + b n + c$ $an^2 + bn + c$ , so in your case a=14, b=24, c=-5004 .

A slightly more complicated way is to reverse foil, trying to factor out your a, b, and c so you can write you equation into
$(x n + y) (s n + t) = 0$ $(xn + y)(sn +t) = 0$ , where x, y, s, and t are all numbers. Then you can solve one $()$ $()$ at a time. But if you have a calculator, the quadratic formula is the best way to go. It'll work every time.

Answer 2 · 2018-03-05T21:52:58

$n = - \frac{6}{7} \pm \frac{15}{7} \sqrt{78}$ $n = -6/7+-15/7sqrt(78)$

Explanation:

Complete the square then use the difference of squares identity to find:

$0 = \frac{7}{2} (14 n^{2} + 24 n - 5004)$ $0 = 7/2(14n^2+24n-5004)$

$0 = 49 n^{2} + 84 n - 17514$ $color(white)(0) = 49n^2+84n-17514$

$0 = {(7 n)}^{2} + 2 (7 n) (6) + 36 - 17550$ $color(white)(0) = (7n)^2+2(7n)(6)+36-17550$

$0 = {(7 n)}^{2} + 2 (7 n) (6) + 6^{2} - (15^{2} \cdot 78)$ $color(white)(0) = (7n)^2+2(7n)(6)+6^2-(15^2 * 78)$

$0 = {(7 n + 6)}^{2} - {(15 \sqrt{78})}^{2}$ $color(white)(0) = (7n+6)^2-(15sqrt(78))^2$

$0 = ((7 n + 6) - 15 \sqrt{78}) ((7 n + 6) + 15 \sqrt{78})$ $color(white)(0) = ((7n+6)-15sqrt(78))((7n+6)+15sqrt(78))$

$0 = (7 n + 6 - 15 \sqrt{78}) (7 n + 6 + 15 \sqrt{78})$ $color(white)(0) = (7n+6-15sqrt(78))(7n+6+15sqrt(78))$

Hence:

$7 n = - 6 \pm 15 \sqrt{78}$ $7n = -6+-15sqrt(78)$

So:

$n = - \frac{6}{7} \pm \frac{15}{7} \sqrt{78}$ $n = -6/7+-15/7sqrt(78)$

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How do you solve $14 n^{2} + 24 n - 5004 = 0$ $14n^2 + 24n - 5004 = 0$ ?

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How do you solve 14n^2 + 24n - 5004 = 014n2+24n−5004=014n^2 + 24n - 5004 = 0?

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How do you solve $14 n^{2} + 24 n - 5004 = 0$ $14n^2 + 24n - 5004 = 0$ ?