Answer:
-1 and 2/3
Step-by-step explanation:
Hello, please consider the following.
[tex]f(x)=f(2x+1)\\\\\begin{aligned}<=> x^2-3x+4&=(2x+1)^2-3(2x+1)+4\\\\&=4x^2+4x+1-6x-3+4\\\\&=4x^2-2x+2\end{aligned}\\\\<=> x^2-3x+4=4x^2-2x+2\\\\<=> 3x^2+x-2=0[/tex]
[tex]\Delta=b^2-4ac=1+4*2*3=25=5^2\\\\x_1=\dfrac{-1-5}{6}=-1\\\\x_2=\dfrac{-1+5}{6}=\dfrac{4}{6}=\dfrac{2}{3}[/tex]
Thank you
