SOLUTION: How do you solve 3y^2=20 using the quadratic formula?

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Question 642162: How do you solve 3y^2=20 using the quadratic formula?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
3y%5E2=20

3y%5E2-20=0

3y%5E2%2B0y-20=0

Now use the quadratic formula to solve for y

In this case, a = 3, b = 0, c = -20,

Plug all this into the quadratic formula to get

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

y+=+%28-%280%29%2B-sqrt%28%280%29%5E2-4%283%29%28-20%29%29%29%2F%282%283%29%29

y+=+%280%2B-sqrt%280-%28-240%29%29%29%2F%286%29

y+=+%28%22%22%2B-sqrt%280%2B240%29%29%2F%286%29

y+=+%28%22%22%2B-sqrt%28240%29%29%2F6

y+=+%28sqrt%28240%29%29%2F6 or y+=+%28-sqrt%28240%29%29%2F6

y+=+%284%2Asqrt%2815%29%29%2F6 or y+=+%28-4%2Asqrt%2815%29%29%2F6

y+=+%282%2Asqrt%2815%29%29%2F3 or y+=+%28-2%2Asqrt%2815%29%29%2F3

y+=+2.58198889747161 or y+=+-2.58198889747161

Note: The last solutions above are approximate