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

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Answer 1 · 2016-10-10T20:42:59

#-3/2# and 2

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#