SOLUTION: solve for y the equation 3(y2+3)=28y.

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Question 826742: solve for y the equation 3(y2+3)=28y.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, please use the "^" character (shift 6) to indicate exponents. Your equation should read:
3(y^2+3)=28y

3%28y%5E2%2B3%29=28y
To solve this we need it in standard form: ay%5E2%2Bby%2Bc+=+0. So we start by transforming the equation. First, simplify:
3y%5E2%2B9=28y
Subtract 28y:
3y%5E2-28y%2B9=0
Next, factor (or use the Quadratic Formula):
%283x-1%29%28x-9%29+=+0
From the Zero Product Property:
3x-1 = 0 or x-9 = 0
Solving these we get:
x+=+1%2F3 or x+=+9