Try to write $n^2 - 2n - 48$ in the form $(n + a)(n + b)$. What must $a$ and $b$ multiply to, and what must they add to?

You will get two values for $n$. After finding them, pay close attention to which one is positive, because the question asks specifically for the positive value of $n$.

In Desmos, type the expression as a function, for example: y = x^2 - 2x - 48. (Using $x$ instead of $n$ is fine; the variable name does not change the solutions.)

Look at where the graph crosses the x-axis (where $y = 0$). You can tap or click on the intersection points to see their coordinates.

Read both x-intercepts from the graph. One will be negative and one will be positive; the positive x-value is the solution the question is asking for.

The equation $n^2 - 2n - 48 = 0$ is a quadratic in standard form $ax^2 + bx + c = 0$. For many SAT questions like this, the fastest method is to factor the quadratic.

To factor $n^2 - 2n - 48$, look for two numbers $a$ and $b$ such that:

Check factor pairs of $-48$: $(1,-48)$, $(2,-24)$, $(3,-16)$, $(4,-12)$, $(6,-8)$, and their reverses. The pair that adds to $-2$ is $6$ and $-8$ because $6 + (-8) = -2$ and $6 \cdot (-8) = -48$.

Use $a = 6$ and $b = -8$ to factor:

Set each factor equal to zero and solve each simple linear equation to obtain two solutions (one negative and one positive).

The two solutions are $n = -6$ and $n = 8$. The problem asks for the positive value of $n$, so we choose $n = 8$, which corresponds to answer choice B) 8.