The question asks for the values of $p$ and $q$ in the quadratic equation:

$x^2 + px + q = 0$

It provides two roots of the equation: $-2$ and $3$.

Using Vieta's formulas for a quadratic equation $x^2 + px + q = 0$, where the roots are $r_1$ and $r_2$:

1. The sum of the roots is equal to $-p$:
   $r_1 + r_2 = -p$ In this case: $-2 + 3 = 1 \implies p = -1$

2. The product of the roots is equal to $q$:
   $r_1 \times r_2 = q$ In this case: $-2 \times 3 = -6 \implies q = -6$

Therefore, the values of $p$ and $q$ are:

$p = -1, \quad q = -6$

Would you like further explanation on how Vieta's formulas work?

Here are 5 questions to expand your understanding:

1. How do Vieta’s formulas apply to higher-degree polynomials?
2. What are the conditions for a quadratic equation to have real roots?
3. How can you derive the quadratic formula from the standard form of a quadratic equation?
4. Can a quadratic equation have complex roots, and how would that affect $p$ and $q$?
5. How do you factor a quadratic equation based on its roots?

Tip: The discriminant $\Delta = b^2 - 4ac$ helps determine the nature of the roots—real or complex—of a quadratic equation.