Recall that the normal distribution $N(\mu, \sigma^2)$ has pdf $f(x\mid \mu ,\sigma ^{2})={\frac {1}{{\sqrt {2\pi \sigma ^{2}}}\ }}\exp {\left(-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}\right)},$ Note here that $\mu = \theta$ and $\sigma^2 = \theta$ and therefore $\sigma = \sqrt{\theta}$
\begin{aligned}
L(x_1,x_2,...,x_n | \theta) &= \prod_{i=1}^n f(x_i | \theta)
\\
&= \prod_{i=1}^n \frac{1}{\sqrt{2 \pi \theta}} \ \exp \Big \{ - \frac{1}{2 \theta} (x_i - \theta)^2 \Big\}
\\
& = (2 \pi)^{-n/2} (\theta)^{-n /2} \prod_{i=1}^n \ \exp \Big \{ - \frac{1}{2 \theta} (x_i - \theta)^2 \Big\}
\\
& = (2 \pi)^{-n/2} (\theta)^{-n /2} \ \exp \Big \{ - \frac{1}{2 \theta} \sum_{i=1}^n (x_i - \theta)^2 \Big\}
\\
\log L& = - \frac{n}{2} \log(2\pi) - \frac{n}{2} \log(\theta) - \frac{1}{2 \theta} \sum_{i=1}^n (x_i - \theta)^2
\end{aligned}
Consider the term $\frac{1}{2 \theta} \sum_{i=1}^n (x_i - \theta)^2$ which can be expanded and simplified
\begin{aligned}
\frac{1}{2 \theta} \sum_{i=1}^n (x_i - \theta)^2 & = \frac{1}{2 \theta} \sum_{i=1}^n (x_i - \theta)(x_i - \theta)
\\
& = \frac{1}{2 \theta} \sum_{i=1}^n \left( x_i^2 - 2 \theta x_i + \theta^2 \right)
\\
& = \frac{1}{2 \theta} \left( \sum_{i=1}^n (x_i^2) - 2 \theta \sum_{i=1}^n (x_i) + n\theta^2 \right)
\\
& = \frac{1}{2 \theta} \sum_{i=1}^n (x_i^2) - \sum_{i=1} (x_i) + \frac{n\theta}{2}
\end{aligned}
We can now compute the derivative with respect to $\theta$, equate to zero and solve for $\theta$
\begin{aligned}
\log L& = - \frac{n}{2} \log(2\pi) - \frac{n}{2} \log(\theta) - \left( \frac{1}{2 \theta} \sum_{i=1}^n (x_i^2) - \sum_{i=1} (x_i) + \frac{n\theta}{2} \right)
\\
\frac{d}{d\theta} \log L & = \frac{-n}{2\theta} - \left( \frac{-1}{2\theta^2} \sum_{i=1}^n (x_i^2) + \frac{n}{2} \right) = 0
\\
& = \frac{-n}{2\theta} + \frac{1}{2\theta^2} \sum_{i=1}^n (x_i^2) - \frac{n}{2}
\\
& = - \theta^2 - \theta + \frac{1}{n} \sum_{i=1}^n (x_i^2)
\\
&\text{let $s = \frac{1}{n} \sum_{i=1}^n (x_i^2)$}
\\
0 & = - \theta^2 - \theta + s
\\
\hat \theta &= \frac{\sqrt{1 + 4s} -1 }{2}
\end{aligned}