1. The problem is to express the equation $x^2 - 3s + 1 = 0$ in a clear form. 2. Notice that the equation involves two variables, $x$ and $s$, but it is quadratic only in $x$. 3. To express $x$ in terms of $s$, we treat $s$ as a constant and solve the quadratic equation for $x$. 4. The quadratic equation is $x^2 - 3s + 1 = 0$. 5. Rearranged, it is $x^2 = 3s - 1$. 6. Taking the square root of both sides, we get $$x = \pm \sqrt{3s - 1}$$. 7. This expression shows $x$ as a function of $s$, valid when $3s - 1 \geq 0$, i.e., $s \geq \frac{1}{3}$. 8. Therefore, the solution set for $x$ is $$x = \pm \sqrt{3s - 1}$$ for $s \geq \frac{1}{3}$.