Step 1: Concept
Treat the equation as a quadratic in $|x|$.
Step 2: Analysis
$a|x|^2+2b|x|-c=0$. Using the quadratic formula: $|x| = \frac{-2b \pm \sqrt{4b^2 + 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 + ac}}{a}$.
Step 3: Evaluation
Since $a, b, c>0$, $\sqrt{b^2 + ac}>b$. Thus, one value of $|x|$ is positive: $|x| = \frac{-b + \sqrt{b^2+ac}}{a}$. The other value is negative and rejected as $|x| \ge 0$.
Step 4: Conclusion
One positive value for $|x|$ yields two real values for $x$ (one positive, one negative).
Final Answer: (b)