Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$.
After much trial and error, I was able to rearrange the above equation
as $80z^2 - 112z-4 = 0$.
$z = \frac{7\pm3\sqrt{6}}{10}$.
I substituted $\frac{7+3\sqrt{6}}{10}$ as $z$ in the $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Things worked out.
Am I still not done? Must I check for extraneous solutions in this situation?
I wonder if it might not require it because if ever we know one of the roots of a quadratic with rational coefficients is irrational, we know the other.
I suppose I'm also trying to avoid going through another round of substitution and test with these irrational conjugates. I'm not very trusting of my algebra yet. So I end up rewriting the same equations time and time again.

$$\sqrt{5 z+5}=2 \sqrt{z}+\sqrt{3-3 z}\tag{1}$$
Square both sides
$$5z+5=4z+3-3z+4\sqrt{3z-3z^2}$$
$$2+4z=4\sqrt{3z-3z^2}\to 1+2z=2\sqrt{3z-3z^2}$$
Square again both sides
$$1+4z+4z^2=12z-12z^2$$
$$16 z^2-8 z+1=0\to (4z-1)^2=0\to z=\frac14$$
Which is actually a solution of $(1)$ (verify!)
