I need to make a correction to your equation. First of all, your notation is off, as @PatrickLI noted. Let's call your mean $E(X)$ and your average of data squared (i.e. $\frac{1}{n} \sum_i x_i^2$) $E(X^2)$, where $X$ is the random variable associated with the above uniform distribution.
Using similar manipulations as you made, I get
$$b^2 - 2 E(X) b -[3 E(X^2)-4E(X)^2]=0$$
which may be solved using the quadratic formula:
$$b=E(X) + \sqrt{3} \sqrt{E(X^2) - E(X)^2}$$
Then $a=2 E(X)-b = E(X) - \sqrt{3} \sqrt{E(X^2) - E(X)^2}$
I chose the $+$ solution for $b$ because $b>a$. I hope this helps.
NOTE
You can verify this solution for $a$ and $b$ with random data in a program that generates uniform random variates such as Mathematica. For example, here is some Mathematica code that generates values of $(a,b)$ from an input interval and a number of data points to generate:
devCheck[left_, right_, n_] := Module[{m1, m2, q},
q = RandomVariate[UniformDistribution[{left, right}], n];
m1 = Mean[q];
m2 = Mean[q^2];
a = m1 - Sqrt[3] Sqrt[m2 - m1^2];
b = m1 + Sqrt[3] Sqrt[m2 - m1^2];
{a, b}];
The question, as you wrote it, is worded in an unclear way. My interpretation is the following:
Suppose you have a sample of $X_i\sim \mathcal{N}(\mu,\sigma)$, which are i.i.d. Find the Maximum likelihood estimate for the two parameters, $\mu$ and $\sigma$.
So
$$\mathcal{L}(\mu,\sigma)=(\frac{1}{\sqrt{2\pi\sigma^2}})^n\cdot \prod_{i=1}^ne^{-\frac{(X_i-\mu)^2}{2\sigma^2}}$$
which is the likelihood function, that we seek to maximize.
To that end, we take its logarithm; it will make the calculation easier, while preserving the extrema, being a monotonically increasing function. Thus
$$\mathcal{F}=\ln(\mathcal{L}(\mu,\sigma))=-\frac{n}{2}\ln(2\pi)-\frac{n}{2} \ln(\sigma^2)+\sum_{i=1}^n-\frac{(X_i-\mu)^2}{2\sigma^2}$$
Now, you simply differentiate with respect to the two parametrs and equate to zero
$$\frac{\partial}{\partial\mu}\mathcal{F}=0\implies \mu=\sum_{i=1}^n\frac{X_i}{n}$$
$$\frac{\partial}{\partial\sigma}\mathcal{F}=0\implies \sigma^2=\sum_{i=1}^n\frac{(X_i-\mu)^2}{n}$$
So, the estimated distribution, given $\bar X=\sum_{i=1}^n\frac{X_i}{n}$, is
$$X\sim \mathcal{N}\left(\bar X,\sqrt{\sum_{i=1}^n\frac{(X_i-\bar X)^2}{n}}\right)$$
Thus it follows that
$$P(X>c)=1-\phi\left(\frac{c-\bar X}{\sqrt{\sum_{i=1}^n\frac{(X_i-\bar X)^2}{n}}}\right)$$
Best Answer
An easier method-of-moments estimator arises from considering the mean. Notice that $E(X) = \frac{a}{2}$, so that one can take $$\hat{a}_{MoM} = 2 \overline{X}$$ as an estimator. Your approach would also work, since we have $$\operatorname{Var}(X) = \frac{a}{12}$$ so that $$a_{MoM}^\prime = 12 \hat{S}^2$$ where $\hat{S}^2$ is the sample variance.