Solving Radical equations

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mcruz65

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Feb 12, 2010
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y=?(3y+10)

Where do I begin? Do I pick numbers for y at random until I get a perfect square?? Suggestions??
 
mcruz65 said:
y=?(3y+10)
Where do I begin? Do I pick numbers for y at random until I get a perfect square?? Suggestions??
Click to expand...
Square both sides: y^2 = 3y + 10
y^2 - 3y - 10 = 0
Can you now solve for y?
 
Yes I can solve it by using the quadratic equation. Results were 5, 2. Thanks alot.
 
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