Question

Решите, пожалуйста, уравнение.

Answer 1

$\displaystyle\\f(x)=3x^2+18x+8\\\\f'(x)-f(x)=0\\\\\\f'(x)=6x+18\\\\6x+18-(3x^2+18x+8)=0\\\\6x+18-3x^2-18x-8=0\\\\-3x^2-12x+10=0\\\\3x^2+12x-10=0\\\\D=12^2-4*3*(-10)=144+12*10=144+120=264\\\\\sqrt{D}=\sqrt{264}=2\sqrt{66}\\\\x=\frac{-12\pm2\sqrt{66}}{6}=\boxed{\frac{-6\pm\sqrt{66}}{3} }$