1. **State the problem:** Solve for all values of $x$ in the equation $$\log_5(4x^2 - 1) - \log_5(4x + 2) = 0.$$\n\n2. **Recall the logarithm property:** The difference of logarithms with the same base can be written as the logarithm of a quotient: $$\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right).$$\n\n3. **Apply the property:** Rewrite the equation as $$\log_5\left(\frac{4x^2 - 1}{4x + 2}\right) = 0.$$\n\n4. **Use the definition of logarithm:** Since $\log_5(M) = 0$ means $M = 5^0 = 1$, we have $$\frac{4x^2 - 1}{4x + 2} = 1.$$\n\n5. **Solve the equation:** Multiply both sides by $4x + 2$ to clear the denominator: $$\cancel{\frac{4x^2 - 1}{4x + 2}} \times (4x + 2) = 1 \times (4x + 2) \Rightarrow 4x^2 - 1 = 4x + 2.$$\n\n6. **Bring all terms to one side:** $$4x^2 - 1 - 4x - 2 = 0 \Rightarrow 4x^2 - 4x - 3 = 0.$$\n\n7. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=4$, $b=-4$, and $c=-3$.\nCalculate the discriminant: $$\Delta = (-4)^2 - 4 \times 4 \times (-3) = 16 + 48 = 64.$$\n\n8. **Find the roots:** $$x = \frac{-(-4) \pm \sqrt{64}}{2 \times 4} = \frac{4 \pm 8}{8}.$$\n\n9. **Calculate each solution:**\n- $$x = \frac{4 + 8}{8} = \frac{12}{8} = \frac{3}{2} = 1.5.$$\n- $$x = \frac{4 - 8}{8} = \frac{-4}{8} = -\frac{1}{2} = -0.5.$$\n\n10. **Check for domain restrictions:** The arguments of the logarithms must be positive:\n- For $\log_5(4x^2 - 1)$, require $4x^2 - 1 > 0 \Rightarrow 4x^2 > 1 \Rightarrow x^2 > \frac{1}{4} \Rightarrow x > \frac{1}{2}$ or $x < -\frac{1}{2}.$\n- For $\log_5(4x + 2)$, require $4x + 2 > 0 \Rightarrow x > -\frac{1}{2}.$\n\n11. **Check each solution:**\n- $x = 1.5$ satisfies $x > \frac{1}{2}$ and $x > -\frac{1}{2}$, so valid.\n- $x = -0.5$ equals $-\frac{1}{2}$, but $4x + 2 = 4(-0.5) + 2 = 0$, which is not allowed (logarithm argument must be > 0). So discard $x = -0.5$.\n\n**Final answer:** $$\boxed{x = \frac{3}{2}}.$$