$a_n$ is an arithmetic progression. We are asked to find the maximum value of $a_1 a_2 - (a_1)^2$ given that its $7th$ term is $2$.
This means that $a + 6d$ = $2$.
From the question, $a + 6d$ = $2$.
I tried taking $a_1 a_2 - (a_1)^2$ as $a(a+d) - a^2]$ to get everything in terms of $a$, which just equals $ad$. I then took $d = \frac{2-a}{6}$, and plugged it into $ad$.
Yet, this doesn't equal anything and the question only states to find its maximum.
How could I go about solving this question?
First observe $$a_1 a_2 - a_1^2 = a_1 (a_2 - a_1) = a_1 d,$$ where $d$ is the common difference. Then $a_7 = a_1 + 6d = 2$ implies $a_1 = 2 - 6d$, hence $$a_1 a_2 - a_1^2 = (2-6d)d = 2d-6d^2.$$ So far, this is equivalent to what you have already done. The crucial missing step is to complete the square:
$$\begin{align} 2d-6d^2 &= -6\left(d^2 - \frac{1}{3}d\right) \\ &= -6\left(d^2 - \frac{1}{3}d + \frac{1}{36} - \frac{1}{36}\right) \\ &= -6 \left(d^2 - 2 \cdot \frac{1}{6}d + \frac{1}{6^2}\right) + \frac{1}{6} \\ &= -6 \left(d - \frac{1}{6}\right)^2 + \frac{1}{6}. \end{align}$$ Because $(d-1/6)^2 \ge 0$ for any real number $d$, the first term is never positive. The maximum value occurs if we can make $-6(d-1/6)^2 = 0$, which is satisfied for $d = 1/6$, and the maximum value attained is $1/6$. This in turn gives us $$a_1 = 1, \quad a_2 = \frac{7}{6}, \quad a_n = 1 + \frac{n-1}{6} = \frac{n+5}{6}.$$