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

1 Answer
Oct 10, 2016

-3/2 and 2

Explanation:

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

2n^2-n-6=0

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

(-b+-sqrt(b^2-4ac))/(2a)

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

(1+-sqrt(1+48))/4

(1+-sqrt49)/4

Take the square root of 49

(1+-7)/4

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

(1+7)/4=8/4=2

So, one of the answers is 2

(1-7)/4=-6/4=-3/2

So, the second answer is -3/2