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

##### 1 Answer
Oct 10, 2016

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

#### Explanation:

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

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

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

$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

So, pug in your numbers into the formula

(1+-sqrt(-1^2-4xx2xx-6))/(2(2)

Solve what's under the radical first

$\frac{1 \pm \sqrt{1 + 48}}{4}$

$\frac{1 \pm \sqrt{49}}{4}$

Take the square root of 49

$\frac{1 \pm 7}{4}$

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

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

So, one of the answers is 2

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

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