SOLUTION: Solve for y in the equation 3(y^2 +3)= 28y

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Question 824355: Solve for y in the equation
3(y^2 +3)= 28y
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
3(y^2+3)= 28y

3y^2+9 = 28y

3y^2+9-28y = 0

3y^2-28y+9 = 0

Use the quadratic formula to solve for y

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

y+=+%28-%28-28%29%2B-sqrt%28%28-28%29%5E2-4%283%29%289%29%29%29%2F%282%283%29%29 Plug in a+=+3, b+=+-28, c+=+9

y+=+%2828%2B-sqrt%28784-%28108%29%29%29%2F%286%29

y+=+%2828%2B-sqrt%28676%29%29%2F6

y+=+%2828%2Bsqrt%28676%29%29%2F6 or y+=+%2828-sqrt%28676%29%29%2F6

y+=+%2828%2B26%29%2F6 or y+=+%2828-26%29%2F6

y+=+54%2F6 or y+=+2%2F6

y+=+9 or y+=+1%2F3


The two solutions are

y+=+9 or y+=+1%2F3