Lagrange Multiplier question

Given a utility function $R(x,y) = x(100-6x) + y(192-4y)$ and a constraint equation $C(x,y) = 2x^2+2y^2+4xy-8x=-20$, maximize.

As usual with Lagrange, I got stuck on the juicier part of solving a set of equations: Lagrangian function $L = x(100-6x) + y(192-4y) - \lambda(2x^2+2y^2+4xy-8x+20)$

$$L_x = 100−12x-λ(4x+4y−8)$$

$$L_y = 192-8y-4\lambda(x+y)$$

setting $L_x$ and $L_y$ to 0, I get $12x+\lambda(4x+4y-8)$ = 100 and $8y-4\lambda(x+y) = 192$

And here's where I got stuck. Anyone have any ideas?

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You should also compute $\frac{dL}{d\lambda}$ and set that to 0. –  tards Nov 13 '11 at 1:45
I'm not getting the same first equation as you. I get $100-12x=\lambda(4x+4y-8)$. –  Grumpy Parsnip Nov 13 '11 at 2:05
Oh whoops, my bad. –  shoopdawoop Nov 13 '11 at 2:22