How do you solve $n^{2} - 17 = 64$ $n^2 - 17=64$ using the quadratic formula?

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How do you solve $n^{2} - 17 = 64$ $n^2 - 17=64$ using the quadratic formula?

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Answer 1 · 2016-08-05T21:02:09

$n = \pm 9$ $n=+-9$

Explanation:

$n^{2} - 17 = 64$ $n^2-17=64$
$n^{2} = 64 + 17$ $n^2=64+17$
$n^{2} = 81$ $n^2=81$
$n = \sqrt{81}$ $n=sqrt81$
$n = \pm 9$ $n=+-9$

Answer 2 · 2016-08-05T21:04:52

$n = \pm 9$ $n=+-9$

Explanation:

Write as $n^{2} - 81$ $n^2-81$

As it is insisted we use the quadratic formula write as:

$n^{2} + 0 n - 81 = 0$ $n^2+0n-81=0$

$\Rightarrow n = \frac{- 0 \pm \sqrt{0^{2} - 4 (1) (- 81)}}{2 (1)}$ $=>n=(-0+-sqrt(0^2-4(1)(-81)))/(2(1))$

$\Rightarrow n = \pm \frac{\sqrt{324}}{2} = \frac{\pm 18}{2} = \pm 9$ $=>n=+-sqrt(324)/2=(+-18)/2 = +-9$

Answer 3 · 2016-08-05T22:06:58

$n = 9, - 9$ $n=9, -9$

Explanation:

$n^{2} - 17 = 64$ $n^2-17=64$

Subtract $64$ $64$ from both sides of the equation.

$n^{2} - 17 - 64 = 0$ $n^2-17-64=0$

Simplify.

$n^{2} - 81$ $n^2-81$

This equation is in the form of a quadratic equation, $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$ , where $a = 1$ $a=1$ , $b = 0$ $b=0$ , and $c = - 81$ $c=-81$ .

The quadratic formula can be used to solve this quadratic equation.

$x = \frac{- b^{2} \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b^2+-sqrt(b^2-4ac))/(2a)$

Substitute $n$ $n$ for $x$ $x$ and plug the known values into the formula.

$n = \frac{- 0 \pm \sqrt{0^{2} - 4 \cdot 1 \cdot - 81}}{2 \cdot 1}$ $n=(-0+-sqrt(0^2-4*1*-81))/(2*1)$

Simplify.

$n = \frac{\pm \sqrt{324}}{2}$ $n=(+-sqrt(324))/(2)$

Simplify.

$n = \pm \frac{18}{2}$ $n=+-18/2$

Simplify.

$n = \pm 9$ $n=+-9$

Solve for $n$ $n$ .

$n = 9$ $n=9$

$n = - 9$ $n=-9$

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How do you solve $n^{2} - 17 = 64$ $n^2 - 17=64$ using the quadratic formula?

3 Answers

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How do you solve n2−17=64n^2 - 17=64 using the quadratic formula?

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How do you solve $n^{2} - 17 = 64$ $n^2 - 17=64$ using the quadratic formula?