Is it Quadratic or not?

$\begin{aligned} x^4-x^2+1 & = k \\ x^4-x^2+1 - k & = 0 \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{This is a quadratic equation of }x^2} \\ \Rightarrow x^2 & = \frac{1 \pm \sqrt{1-4+4k}}{2}\\ & = \frac{1 \pm \sqrt{4k-3}}{2} \end{aligned}$

For real roots, $\begin{cases} 4k-3 \ge 0 & \implies k \ge \dfrac{3}{4} \\ \dfrac{1 - \sqrt{4k-3}}{2} \ge 0 & \implies 1 - \sqrt{4k-3} \ge 0 & \implies \sqrt{4k-3} \le 1 & \implies 4k-3 \le 1 & \implies k \le 1 \end{cases}$

$\begin{aligned} \implies \frac{3}{4} \le k & \le 1 \implies 6 \le 8k & \le 8 \implies \end{aligned}$ There are $\boxed{3}$ integer values of $8k$ , which are $6$ , $7$ and $8$ .