How do you solve the quadratic using the quadratic formula given $2 n^{2} - n - 4 = 2$ $2n^2-n-4=2$ ?

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How do you solve the quadratic using the quadratic formula given $2 n^{2} - n - 4 = 2$ $2n^2-n-4=2$ ?

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Answer 1 · 2016-10-10T20:42:59

$- \frac{3}{2}$ $-3/2$ and 2

Explanation:

First, you have to make the equation equal 0, so subtract 2 on both sides of the equation

$2 n^{2} - n - 6 = 0$ $2n^2-n-6=0$

Next, you are going to plug numbers into the quadratic equation, which is:

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

In your equation, the $a, b, and c$ $a,b, and c$ are in order of your equation. So, your $a = 2$ $a=2$ from $2 n$ $2n$ , $b = - 1$ $b=-1$ from $- n$ $-n$ , and $c = - 6$ $c=-6$ from $- 6$ $-6$

So, pug in your numbers into the formula

$\frac{1 \pm \sqrt{- 1^{2} - 4 \times 2 \times - 6}}{2 (2)}$ $(1+-sqrt(-1^2-4xx2xx-6))/(2(2)$

Solve what's under the radical first

$\frac{1 \pm \sqrt{1 + 48}}{4}$ $(1+-sqrt(1+48))/4$

$\frac{1 \pm \sqrt{49}}{4}$ $(1+-sqrt49)/4$

Take the square root of 49

$\frac{1 \pm 7}{4}$ $(1+-7)/4$

Because the problem was a square root, and there is a $\pm$ $+-$ , there will be 2 answers.

$\frac{1 + 7}{4} = \frac{8}{4} = 2$ $(1+7)/4=8/4=2$

So, one of the answers is 2

$\frac{1 - 7}{4} = - \frac{6}{4} = - \frac{3}{2}$ $(1-7)/4=-6/4=-3/2$

So, the second answer is $- \frac{3}{2}$ $-3/2$

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How do you solve the quadratic using the quadratic formula given $2 n^{2} - n - 4 = 2$ $2n^2-n-4=2$ ?

1 Answer

Explanation:

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How do you solve the quadratic using the quadratic formula given $2 n^{2} - n - 4 = 2$ $2n^2-n-4=2$ ?