[tex]\displaystyle\bf\\f(x)=5\Big(2x^{2} -8x+12\Big)-6=10x^{2} -40x+60-6=10x^{2} -40x+54\\\\\\f'(x)=10\cdot(x^{2} )'-40\cdot x'+54'=10\cdot 2x-40\cdot 1+0=20x-40\\\\\\f'(x)=0 \ \ \ \Rightarrow \ \ \ 20x-40=0\\\\\\20x=40\\\\\\\boxed{\boxed{x=2}}[/tex]