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I'm trying to solve the following for $z_1$, $z_2$, and $t$:

$$tCz_2^2-tD-\epsilon_2z_2=0$$ $$tAz_1^2-tB-\delta_2z_1=0$$ $$\epsilon_1Az_1^2z_2+\epsilon_1Bz_2+C\delta_1z_1z_2^2+D\delta_1z_1+(-2\delta_1+E-2\epsilon_1)z_1z_2=0$$ Where $A, B, C, D, E, \epsilon_1, \epsilon_2, \delta_1, \delta_2$ are constants.

I've tried using polar coordinates ($z_1=r\cos\theta$, $z_2=r\sin\theta$) then solve them for specific values of $\theta$ then find the $t$ and $r$ accordingly e.g., when $\theta=(2n+1)\pi,\theta=2n\pi, \theta=(2n+1/2)\pi$, and $\theta=(2n+3/2)\pi$. But I wonder if there is an explicit solution to the above system of equations?

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    The last equation is of (total) degree 3 [not quadratic as title of post suggests] – 2017-02-24
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    sorry, I mean't quadratic in each variable. – 2017-02-24
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    Any luck solving first two for $t$ and setting results equal? (at least gets the system down to two variables...) – 2017-02-24
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    @LewisProctor You get a quartic in $t$ by eliminating $z_1,z_2$ between the equations (using resultants for example), but the calculations are not pretty, and it's unlikely to get any "nice" solution unless there are additional relations between those constants. – 2017-02-24
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    I'm actually doing that now. I solved for $t$, then reduced it to two simultaneous cubic equations. – 2017-02-24

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