If \(f(x)=x^{2}-3 x+4\) and \(f(x)=f(2 x+1)\), then \(x=\)

\(\text { Here } \mathrm{f}(2 \mathrm{x}+1\)
\(=(2 \mathrm{x}+1)^{2}-3(2 \mathrm{x}+1)+4 \)
\( =4 \mathrm{x}^{2}+4 \mathrm{x}+1-6 \mathrm{x}-3+4 \)
\( \therefore \mathrm{f}(2 \mathrm{x}+1) =4 \mathrm{x}^{2}-2 \mathrm{x}+2 \)
\( \text {Given} \mathrm{f}(\mathrm{x})=\mathrm{f}(2 \mathrm{x}+1) \)
\( \therefore \mathrm{x}^{2}-3 \mathrm{x}+4=4 \mathrm{x}^{2}-2 \mathrm{x}+2 \)
\( \therefore 3 \mathrm{x}^{2}+\mathrm{x}-2 \quad=0 \Rightarrow 3 \mathrm{x}^{2}+3 \mathrm{x}-2 \mathrm{x}-2=0 \)
\(3 \mathrm{x}(\mathrm{x}+1)-2(\mathrm{x}+1)=0 \Rightarrow(\mathrm{x}+1)(3 \mathrm{x}-2)=0 \)
\(\therefore \mathrm{x}=-1, \frac{2}{3}\)