The function \( \mathbf{f} \) is such that \( f(x)=\frac{1}{x+2}+\frac{1}{x-3} \). Given that \( \mathbf{f}(\mathbf{x})=\mathbf{4} \), find the possible values of \( \mathbf{x} \). Let your answer be in the form \( \frac{p+\sqrt{q}}{r} \) where \( \mathbf{p}, \mathbf{q} \) and r are positive integers.

To solve the equation \( f(x) = 4 \), we start with \[ \frac{1}{x+2} + \frac{1}{x-3} = 4. \] First, let's combine the fractions on the left side: \[ \frac{(x-3) + (x+2)}{(x+2)(x-3)} = 4. \] This simplifies to: \[ \frac{2x - 1}{(x+2)(x-3)} = 4. \] Now, we cross-multiply to eliminate the fraction: \[ 2x - 1 = 4(x+2)(x-3). \] Now, let's expand the right side: \[ (x+2)(x-3) = x^2 - 3x + 2x - 6 = x^2 - x - 6. \] Therefore, \[ 4(x^2 - x - 6) = 4x^2 - 4x - 24. \] Setting the equation gives: \[ 2x - 1 = 4x^2 - 4x - 24. \] Bring all terms to one side: \[ 0 = 4x^2 - 4x - 24 - 2x + 1. \] Combining like terms: \[ 0 = 4x^2 - 6x - 23. \] Now, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -6 \), and \( c = -23 \). Calculate the discriminant: \[ b^2 - 4ac = (-6)^2 - 4(4)(-23) = 36 + 368 = 404. \] Now substituting back into the quadratic formula: \[ x = \frac{-(-6) \pm \sqrt{404}}{2 \cdot 4} = \frac{6 \pm \sqrt{404}}{8} = \frac{6 \pm 2\sqrt{101}}{8} = \frac{3 \pm \sqrt{101}}{4}. \] Thus, the possible values of \( x \) are: \[ x = \frac{3 + \sqrt{101}}{4} \quad \text{and} \quad x = \frac{3 - \sqrt{101}}{4}. \] In the form \( \frac{p+\sqrt{q}}{r} \), the values are: \[ \mathbf{p} = 3, \quad \mathbf{q} = 101, \quad \mathbf{r} = 4. \]