SOLUTION: Find the reals of the solution: 2=y+6y^2

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Question 643590: Find the reals of the solution: 2=y+6y^2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
2=y%2B6y%5E2
To solve a quadratic equation, you want one side to be zero. Subtracting 2 from each side:
0=y%2B6y%5E2-2
Then we'll put the terms in "proper" order (highest exponent to lowest):
0=6y%5E2%2By-2
Now we factor:
0+=+%283y%2B2%29%282y-1%29
Next we use the Zero Product Property which tells us that for any product, like this one, to be zero one of its factors must be zero. So:
3y%2B2+=+0 or 2y-1=0
Solving these...
3y+=+-2 or 2y=1
y+=+-2%2F3 or y=1%2F2
Both of these are real numbers.