To solve the problem, let's analyze it step by step.

We have two functions:

1. $f(x) = 2x + 3$
2. $g(x) = x^2 - x + 2$

The equation $g(a) = 4$ means that when we substitute $x = a$ into the function $g(x)$, the output should be 4. So we can write:

$g(a) = a^2 - a + 2 = 4$

Now, solve for $a$:

$a^2 - a + 2 = 4$ $a^2 - a + 2 - 4 = 0$ $a^2 - a - 2 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

$a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$ $a = \frac{1 \pm \sqrt{1 + 8}}{2}$ $a = \frac{1 \pm \sqrt{9}}{2}$ $a = \frac{1 \pm 3}{2}$

So, the two possible solutions are:

$a = \frac{1 + 3}{2} = 2 \quad \text{or} \quad a = \frac{1 - 3}{2} = -1$

Thus, $a = 2$ or $a = -1$.

If you want to explore more, let me know! Here are five related questions:

1. What is $f(2)$ and $f(-1)$?
2. How do we find the derivative of $g(x)$?
3. What happens if $g(a) = 6$? How does that change the result?
4. Can we factor $a^2 - a - 2$ without using the quadratic formula?
5. What is the domain and range of $f(x)$ and $g(x)$?

Tip: Always check if a quadratic can be factored before using the quadratic formula—it can save time!