A number theory problem by Linkin Duck

Let $f(x) = 4x^2 - 30x - 1975,$ let $\lambda$ denote the number of decimal digits of $x$ , and let $p$ denote the product of the digits of $x$ . Then clearly we have $9^\lambda \geq p. \tag{1}$ Suppose $\lambda \geq 3$ so that $x \geq 10^{\lambda - 1} \geq 100$ . Since $f$ is a parabola and is increasing for $x > 30/(2 \cdot 4) = 3.75$ , we also have $x \geq 10^{\lambda - 1} \implies f(x) \geq f(10^{\lambda - 1}). \tag{2}$ Observe that, for all $\lambda \geq 3$ , $\frac 4{100} 10^\lambda - 4 \geq 1.975 \implies 10^\lambda \left( \frac 4{100} 10^\lambda - 4 \right) \geq 10^3 \cdot 1.975 \implies 10^\lambda \left( \frac 4{100} 10^\lambda - 4 \right) - 1975 \geq 0.$ Therefore, for all $\lambda \geq 3$ , $10^\lambda + 10^\lambda \left( \frac 4{100} 10^\lambda - 4 \right) - 1975 \geq 10^\lambda > 9^\lambda. \tag{3}$ The left side of the last inequality is equivalent to $f(10^{\lambda - 1})$ , so $(1)$ , $(2)$ , and $(3)$ together show that, when $\lambda \geq 3$ , $f(x) \geq f(10^{\lambda - 1}) > 9^\lambda \geq p.$ This means that $f(x) = p$ only if $\lambda \leq 2$ . But we can confirm case by case that $f(x) \neq p$ for all single- and double-digit positive integers $x$ . Therefore, there are $\boxed{0}$ positive integers $x$ for which $f(x) = p$ .