14. Given the sequence $\left\{x_{n} \mid\right.$ satisfies $x_{0}=0, x_{n+1}=x_{n}$ $+a+\sqrt{b^{2}+4 a x_{n}}(n=0,1,2, \cdots)$ where $a, b$ are given positive real numbers, find the general term formula for $x_{n}$:
14. $x_{n}=a n^{2}+b n$

Let $y_{n}-\sqrt{b^{2}+4 a x_{n}}$, rearranging gives $x_{n}=\frac{y_{n}^{2}-b^{2}}{4 a}$, and $\frac{y_{n+1}^{2}-b^{2}}{4 a}=x_{n+1}=\frac{y_{n}^{2}-b^{2}}{4 a}+a+y_{n}$
Simplifying, we get $y_{n+1}^{2}=y_{n}^{2}+4 a^{2}+4 a y_{n}, y_{n+1}^{2}=\left(y_{n}+2 a\right)^{2}$ $y_{n+1}=y_{n}+2 a$, which means $\left\{y_{n}\right\}$ is an arithmetic sequence.
Also, $y_{0}=b$, then $y_{n}=b+2 a n$.
$$
x_{n}=\frac{y_{n}^{2}-b^{2}}{4 a}=a n^{2}+b n .
$$