SOLUTION: How do I solve this 2y^2-y=6 quadratic equation using the Quadratic Formula.

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Question 110997: How do I solve this 2y^2-y=6 quadratic equation using the Quadratic Formula.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
+2y%5E2-y=6 Start with the given equation

+2y%5E2-y-6=0 Subtract 6 from both sides


Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for y:


Starting with the general quadratic


ay%5E2%2Bby%2Bc=0


the general solution using the quadratic equation is:


y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 2%2Ay%5E2-y-6=0 ( notice a=2, b=-1, and c=-6)





y+=+%28--1+%2B-+sqrt%28+%28-1%29%5E2-4%2A2%2A-6+%29%29%2F%282%2A2%29 Plug in a=2, b=-1, and c=-6




y+=+%281+%2B-+sqrt%28+%28-1%29%5E2-4%2A2%2A-6+%29%29%2F%282%2A2%29 Negate -1 to get 1




y+=+%281+%2B-+sqrt%28+1-4%2A2%2A-6+%29%29%2F%282%2A2%29 Square -1 to get 1 (note: remember when you square -1, you must square the negative as well. This is because %28-1%29%5E2=-1%2A-1=1.)




y+=+%281+%2B-+sqrt%28+1%2B48+%29%29%2F%282%2A2%29 Multiply -4%2A-6%2A2 to get 48




y+=+%281+%2B-+sqrt%28+49+%29%29%2F%282%2A2%29 Combine like terms in the radicand (everything under the square root)




y+=+%281+%2B-+7%29%2F%282%2A2%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




y+=+%281+%2B-+7%29%2F4 Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


y+=+%281+%2B+7%29%2F4 or y+=+%281+-+7%29%2F4


Lets look at the first part:


x=%281+%2B+7%29%2F4


y=8%2F4 Add the terms in the numerator

y=2 Divide


So one answer is

y=2




Now lets look at the second part:


x=%281+-+7%29%2F4


y=-6%2F4 Subtract the terms in the numerator

y=-3%2F2 Divide


So another answer is

y=-3%2F2


So our solutions are:

y=2 or y=-3%2F2