1. In the set of real numbers, solve the equation

$$
\sqrt{3 x-a}=a-2 x
$$
Solution. If we square the given equation, we get:

$$
4 x^{2}-(4 a+3) x+a^{2}+a=0
$$

Thus,

$$
x_{1 / 2}=\frac{4 a+3 \pm \sqrt{8 a+9}}{8}
$$

Let's check if the found solutions satisfy the original equation. The right side has the value

$$
a-2 x=\frac{-3 \mp \sqrt{8 a+9}}{8} \text{. }
$$

Since $a-2 x$ is a non-negative number, we get that the root can only be

$$
x_{1}=\frac{4 a+3-\sqrt{8 a+9}}{8}
$$

Moreover, from (1) it is clear that $a \geq 0$, so:

$$
\begin{aligned}
a-2 x & =\frac{-3+\sqrt{8 a+9}}{4}=\sqrt{\left(\frac{-3+\sqrt{8 a+9}}{4}\right)^{2}}=\sqrt{\frac{9+4 a-6 \sqrt{8 a+9}}{8}} \\
& =\sqrt{\frac{12 a+9-6 \sqrt{8 a+9}}{8}-a}=\sqrt{3 x-a}
\end{aligned}
$$