[tex]4x^2-3x+9=2x+1 \\
4x^2-3x-2x+9-1=0 \\
4x^2-5x+8=0 \\ \\
a=4 \\ b=-5 \\ c=8 \\ b^2-4ac=(-5)^2 - 4 \times 4 \times 8=25-128=-103 \\ \\
x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-5) \pm \sqrt{-103}}{2 \times 4}=\boxed{\frac{5 \pm \sqrt{103}i}{8}}[/tex]
The answer is C.
