If f(x)=\acx−32x+1f(x) = \ac{x-3}{2x+1}f(x)=\acx−32x+1 (assume xeq−0.5x eq -0.5xeq−0.5), find the value of xxx such that f(x)=0f(x) = 0f(x)=0.
3
-0.5
0
No solution