Confidence interval for Poisson distribution using CLT.

In my lecture notes it is used that if $$x_1,...,x_n$$ is an i.i.d. sample from a Poisson distribution with parameter $$\lambda$$ then a confidence interval for $$\lambda$$ is given as $$\left[ \bar{x}+\frac{q^2}{n}-\frac{q}{\sqrt{n}}\sqrt{\bar{x}-\frac{q^2}{4n}},\bar{x}+\frac{q^2}{n}+\frac{q}{\sqrt{n}}\sqrt{\bar{x}+\frac{q^2}{4n}}\right]$$ where $$\bar{x}=\sum_{i=1}^nx_i$$ and $$q$$ is the $$1-\alpha/2$$ quantile of a standard normal distribution.

So I tried to to show that myself but got stuck. Here is what I got: by CLT for large $$n$$ we have $$\frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}$$is approximately standard normal. Then using that $$P\left(q_{\alpha/2}\le \frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}\le q\right)$$ is a approximately $$1-\alpha$$ and the MLE for $$\lambda$$ is $$\hat{\lambda}=1/\bar{x}$$ I get a interval of the form $$\bar{x}\pm q\sqrt{\hat{\lambda}/n}.$$ However I can't seem to arrive at the interval given in the lecture notes and some help would be appreciated.

We can rewrite the event $$\left\{q_{\alpha/2}\leqslant \frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}\leqslant q\right\}$$ as $$\left\{q_{\alpha/2}\sqrt \lambda\leqslant \sqrt n(\bar{x} - \lambda)\leqslant q\sqrt{\lambda}\right\}$$.
Solving the inequations $$q_{\alpha/2}x\leqslant \sqrt{n}(\overline{x}-x^2)$$ and $$\sqrt{n}(\overline{x}-x^2)\leqslant qx$$ will give that $$x$$ belongs to some interval; then replace $$x$$ by $$\sqrt \lambda$$ to get an interval for $$\lambda$$.