Answer: C. Exactly two

Explanation:1. Analyze the given equation: The equation is $x^{2}-\dfrac{25}{81}=0$. This is a quadratic equation.

2. Solve for $x^2$: Add $\dfrac{25}{81}$ to both sides of the equation: $x^2 = \dfrac{25}{81}$

3. Solve for x by taking the square root of both sides: When taking the square root to solve an equation of the form $x^2 = k$ (where $k>0$), there are two real solutions: $x = \sqrt{k}$ and $x = -\sqrt{k}$. $x = \pm\sqrt{\dfrac{25}{81}}$ $x = \pm\dfrac{\sqrt{25}}{\sqrt{81}}$ $x = \pm\dfrac{5}{9}$

4. Identify the distinct real solutions: The two distinct real solutions are $x\_1 = \dfrac{5}{9}$ and $x\_2 = -\dfrac{5}{9}$.

5. Determine the number of distinct real solutions: There are exactly two distinct real solutions.

6. Alternative method using the discriminant: For a quadratic equation $Ax^2+Bx+C=0$, the discriminant is $D = B^2-4AC$. - If $D>0$, there are two distinct real solutions. - If $D=0$, there is exactly one real solution (a repeated root). - If $D<0$, there are no real solutions (two complex conjugate solutions). The given equation is $x^2 + 0x - \dfrac{25}{81} = 0$. So, $A=1$, $B=0$, $C=-\dfrac{25}{81}$. $D = (0)^2 - 4(1)(-\dfrac{25}{81})$ $D = 0 - (-\dfrac{100}{81})$ $D = \dfrac{100}{81}$ Since $D = \dfrac{100}{81} > 0$, there are exactly two distinct real solutions.

7. Check the solution with the options: Option [C] states "Exactly two", which matches our finding.

8. Conclusion: The given equation has exactly two distinct real solutions.