321. $\sqrt{2 x+5}+\sqrt{x-1}=8$.

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321. $\sqrt{2 x+5}+\sqrt{x-1}=8$.
Solution. Isolate one of the radicals and square both sides of the equation:

$$
(\sqrt{2 x+5})^{2}=(8-\sqrt{x-1})^{2} ; 2 x+5=64-16 \sqrt{x-1}+(x-1)
$$

Move $16 \sqrt{x-1}$ to the left side, and all other terms to the right side:

$$
16 \sqrt{x-1}=64+x-1-2 x-5 ; 16 \sqrt{x-1}=58-x
$$

Square both sides of the equation again:

$(16 \sqrt{x-1})^{2}=(58-x)^{2} ; 256 x-256=3364-116 x+x^{2} ; x^{2}-372 x+3620=0$.

Solve the resulting quadratic equation. Since $a=1, b=-372$, $c=3620, \quad D=b^{2}-4 a c=(-372)^{2}-4 \cdot 1 \cdot 3620=138384-14480=123904$; $\sqrt{D}=352$, then

$$
x_{1}=\frac{-b-\sqrt{D}}{2 a}=\frac{372-352}{2}=10 ; \quad x_{2}=\frac{-b+\sqrt{D}}{2 a}=\frac{372+352}{2}=362
$$

Check both values $x_{1}=10 ; x_{2}=362$.

If $x=10$, then $\sqrt{2 \cdot 10+5}+\sqrt{10-1}=8 ; \sqrt{25}+\sqrt{9}=8 ; 5+3=8$; $8=8$

If $x=362$, then $\sqrt{2 \cdot 362+5}+\sqrt{362-1}=8 ; \sqrt{729}+\sqrt{361}=8 ; 27+$ $+19 \neq 8$; therefore, $x=362$ is an extraneous root.

The root of the given equation is $x=10$.