10.3. Solve in $\mathbb{R}$ the equation $\sqrt{8 x^{2}+10 x-3}-\sqrt{8 x+12}=3+\sqrt{4 x+8}-\sqrt{4 x^{2}+7 x-2}$.
## Solution.

The equation from the statement is equivalent to the equation $\sqrt{(4 x-1)(2 x+3)}+\sqrt{(4 x-1)(x+2)}-2 \sqrt{2 x+3}-2 \sqrt{x+2}=3$.

We will have $D V A=\left[\frac{1}{4} ;+\infty\right)$.

Next, we write the equation in the form

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

Since $\sqrt{2 x+3}+\sqrt{x+2}>0$ for any $x \in D V A$, it follows from (1) that $\sqrt{4 x-1}-2>0$, i.e., $x>\frac{5}{4}$. Therefore, the equation from the statement can have solutions only on the interval $\left(\frac{5}{4} ;+\infty\right)$.

The sum of two strictly increasing functions is a strictly increasing function. Therefore, the function $f(x)=\sqrt{2 x+3}+\sqrt{x+2}$ is strictly increasing on $\left(\frac{5}{4} ;+\infty\right)$, and it is also positive. The function $g(x)=\sqrt{4 x-1}-2$ is also a strictly increasing and positive function on $\left(\frac{5}{4} ;+\infty\right)$.

The product of two positive and strictly increasing functions is a strictly increasing function, which represents the left-hand side of equation (1), while the right-hand side of equation (1) is a constant. Therefore, equation (1) has at most one solution on the interval $\left(\frac{5}{4} ;+\infty\right)$.

We observe that for $x=2 \in\left(\frac{5}{4} ;+\infty\right)$ in (1) we have $(\sqrt{7}-2)(\sqrt{7}+2)=3$, which is true.

Thus, the only solution is $x=2$.

Answer: $S=\{2\}$.