1. **State the problem:** We need to find the possible values of the parameter $r$ such that the function $$h(x) = \log_{10}(4x^2 - rx + r - 1)$$ is defined for all real numbers $x$ (i.e., the argument of the logarithm is positive for all $x \in \mathbb{R}$). 2. **Analyze the domain:** The logarithm function is defined only for positive arguments. So, we require $$4x^2 - rx + r - 1 > 0$$ for all real $x$. 3. **Consider the quadratic expression:** Let $$f(x) = 4x^2 - rx + r - 1.$$ For $f(x) > 0$ for all real $x$, two conditions must hold: - The quadratic opens upward, so the leading coefficient must be positive: $$4 > 0,$$ which is true. - The quadratic equation $f(x) = 0$ has no real roots. So, the discriminant must be negative: $$\Delta = b^2 - 4ac < 0,$$ where $a=4$, $b=-r$, and $c=r-1$. 4. **Calculate the discriminant:** $$\Delta = (-r)^2 - 4 \cdot 4 \cdot (r-1) = r^2 - 16(r-1) = r^2 - 16r + 16.$$ We require $$r^2 -16r + 16 < 0.$$ 5. **Solve the inequality:** The quadratic can be rewritten as $$r^2 - 16r + 16 = 0.$$ The roots are found by the quadratic formula: $$r = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 16}}{2} = \frac{16 \pm \sqrt{256 - 64}}{2} = \frac{16 \pm \sqrt{192}}{2} = \frac{16 \pm 8\sqrt{3}}{2} = 8 \pm 4\sqrt{3}.$$ Since the parabola opens upward, the inequality $r^2 -16r + 16 < 0$ holds between the roots: $$8 - 4\sqrt{3} < r < 8 + 4\sqrt{3}.$$ 6. **Final answer:** The possible values of $r$ are $$\boxed{8 - 4\sqrt{3} < r < 8 + 4\sqrt{3}}.$$