<p>Equation: x=(y^2)+2y-4</p>
<p>How do you solve for “y” ?</p>
<p>This shouldn’t be in the ‘Brown University’ forum…</p>
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<p>True, but we are gracious folks here. First turn the righthand side into a perfect square by adding 5 to both sides: x + 5 = (y^2) + 2y - 4 + 5 = (y^2) + 2y + 1 = (y+1)^2. Now we have x + 5 = (y+1)^2. Take the square root of each side: sqrt{x+5} = y+1 so y = sqrt{x+5} - 1.</p>
<p>Thank you so much! And yeah, I know this forum isn’t exactly for questions like this, but I knew that smart people roam these boards :)</p>
<p>Coase, don’t forget it’s +/- sqrt{x+5} - 1.</p>
<p>Completing the square is quicker, but that might take an intellectual realization that wouldn’t necessarily come to everyone immediately (it didn’t to me until you mentioned it). I would’ve just brought the x over to the right side and treated the (-4-x) as a constant. Then I would’ve went quadratic formula and it simplifies pretty easily there.</p>