Solve for area of A1 when volume and A2 are known for a frustum pyramid

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I'm wondering how to solve out the equation so that A1 is on the left side of the equation when the volume and A2 are known. Any thoughts?

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HINT: Here are steps to remove the square root sign

$V=\frac{h}{3}(A_1+A_2+\sqrt{A_1A_2})$

$\frac{3V}{h}=(A_1+A_2+\sqrt{A_1A_2})$

$\frac{3V}{h}-A_1-A_2=\sqrt{A_1A_2}$

$(\frac{3V}{h}-A_1-A_2)^2=A_1A_2$

$(\frac{3V}{h})^2+A_1^2+A_2^2+2(A_1A_2+\frac{3V}{h}A_1+\frac{3V}{h}A_2)=A_1A_2$

The above is a quadratic equation about $A_1$ and hope you can continue now.