Real quadratic

Upon examination, the value $a = -4$ in the denominator has no real zeros due to $\large x = \frac{-3 \pm \sqrt{9-4(-4)(-4)}}{-8} = \frac{3 \mp \sqrt{55}i}{8}$ , which reduces the rational function $\large f(x) = \frac{ax^2+3x-4}{-4x^2+3x+a}$ to $1$ for all $x \in \mathbb{R}.$

The other way $f(x): \mathbb{R} \rightarrow \mathbb{R}$ remains valid occurs iff $f'(x) \neq 0$ over $x \in \mathbb{R}.$ Its derivative computes to:

$\large f'(x) = \frac{(2ax+3)(-4x^2+3x+a)-(-8x+3)(ax^2+3x-4)}{(-4x^2+3x+a)^2} = \frac{(3a+12)x^2+2(a^2-16)x+(3a+12)}{(-4x^2+3x+a)^2} \neq 0$ ;

which will be satisfied when $(3a+12)x^2+2(a^2-16)x+(3a+12) \neq 0$ , or $\large x = \frac{-2(a^2-16) \pm \sqrt{4(a^2-16)^2 - 4(3a+12)^2}}{2(3a+12)}$ ;

or $\large x = \frac{(16-a^2) \pm \sqrt{(a+4)^2(a-4)^2 -9(a+4)^2}}{3a+12};$

or $\large x = \frac{(4+a)(4-a) \pm (4+a)\sqrt{(a-4)^2 - 9}}{3(a+4)};$

or $\large x = \frac{(4-a) \pm \sqrt{a^2 -8a + 7}}{3};$

or $\large x = \frac{(4-a) \pm \sqrt{(a-1)(a-7)}}{3};$

which requires $(a-1)(a-7) \le 0 \Rightarrow 1 \le a \le 7$ . Hence, the result is $a \in [1,7] \cup \{-4\} \Rightarrow 1 + 7 - 4 = \boxed{4}.$