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!

This problem explains how to solve the quadratic equation g(a) = a² - a + 2 = 4 using the quadratic formula. Follow a step-by-step guide to finding the value of 'a'. Key concepts include solving quadratic equations and applying the quadratic formula. Suitable for Grades 8-10.

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