320. $\sqrt{x-1} \cdot \sqrt{2 x+6}=x+3$.

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320. $\sqrt{x-1} \cdot \sqrt{2 x+6}=x+3$.
Solution. First, we perform the multiplication of the roots:

$\sqrt{(x-1)(2 x+6)}=x+3 ; \sqrt{2 x^{2}-2 x+6 x-6}=x+3 ; \sqrt{2 x^{2}+4 x-6}=x+3$.

Now, we square both sides of the equation:

$\left(\sqrt{2 x^{2}+4 x-6}\right)^{2}=(x+3)^{2} ; 2 x^{2}+4 x-6=x^{2}+6 x+9 ; x^{2}-2 x-15=0$.

We solve the quadratic equation. Here $a=1, b=-2, c=-15, D=$ $=b^{2}-4 a c=4-4 \cdot 1 \cdot(-15)=64, \sqrt{D}=8$; therefore,

$$
x_{1,2}=\frac{-b \pm \sqrt{D}}{2 a} ; x_{1}=\frac{2-8}{2}=-3 ; x_{2}=\frac{2+8}{2}=5 ; x_{1}=-3, x_{2}=5
$$

Let's check both roots.

For $x=-3$ we have $\sqrt{(-3)-1} \cdot \sqrt{2(-3)+6}=-3+3$. We see that the first radical has no meaning in the set of real numbers; therefore, $x=-3$ is an extraneous root.

For $x=5$ we have $\sqrt{5-1} \cdot \sqrt{5 \cdot 2+6}=5+3 ; \sqrt{4} \cdot \sqrt{16}=8 ; 2 \cdot 4=8$. Thus, the root of the given equation is $x=5$.