\(2x-3x^2+4\)
\(=-3\left(x^2-\frac{3}{2}x-\frac{4}{3}\right)\)
\(=-3\left(x^2-2.x.\frac{3}{4}+\frac{9}{16}-\frac{91}{48}\right)\)
\(=\frac{91}{16}-3\left(x^2-\frac{3}{4}\right)^2\le\frac{91}{16}\)
Max = \(\frac{91}{16}\Leftrightarrow x^2-\frac{3}{4}=0\Rightarrow x^2=\frac{3}{4}\Rightarrow x=\sqrt{\frac{3}{4}}\)