SOLUTION: Use the quadratic formula to solve the equation for x in terms of y.? 2x^2 + 9xy -5 - y^2 = 0 This is what I did: (2)x^2 + (9y)x + (-y^2-5) = 0 -9y +/- square root: 9y^2 -4(2

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Use the quadratic formula to solve the equation for x in terms of y.? 2x^2 + 9xy -5 - y^2 = 0 This is what I did: (2)x^2 + (9y)x + (-y^2-5) = 0 -9y +/- square root: 9y^2 -4(2      Log On


   

Question 342241: Use the quadratic formula to solve the equation for x in terms of y.?
2x^2 + 9xy -5 - y^2 = 0
This is what I did:
(2)x^2 + (9y)x + (-y^2-5) = 0
-9y +/- square root: 9y^2 -4(2)(-y^2-5) / 4
-9y +/- square root: 9y^2 + 8y^2 +40/ 4
-9y +/- square root : 17y^2 +40/ 4
Why is this answer wrong? Please explain. Thanks!!
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-9y+%2B-+sqrt%28+%289y%29%5E2-4%2A2%2A%28-%285%2By%5E2%29%29%29%29+%2F%282%2A2%29+
You didn't square the 9y term correctly.
x+=+%28-9y+%2B-+sqrt%28+81y%5E2%2B%2840%2B8y%5E2%29%29%29+%2F4+
x+=+%28-9y+%2B-+sqrt%28+89y%5E2%2B40%29%29+%2F4+