Odpowiedź:
[tex]\huge\boxed{\boxed{x\in\{-4;0;3\}}}[/tex]
Szczegółowe wyjaśnienie:
[tex]2x^3+2x^2=24x\\\\2x^3+2x^2-24x=0 \ \ /:2\\\\x^3+x^2-12x=0\\\\x(x^2+x-12)=0\\\\x_1=0 \ \ \ \vee \ \ \ x^2+x-12=0\\\\a=1, \ b=1, \ c=-12\\\\\Delta=b^2-4ac\Rightarrow1^2-4\cdot1\cdot(-12)=1+48=49\\\\\sqrt{\Delta}=\sqrt{49}=7\\\\x_2=\frac{-b-\sqrt{\Delta}}{2a}\Rightarrow\frac{-1-7}{2\cdot1}=\frac{-8}{2}=-4\\\\x_3=\frac{-b+\sqrt{\Delta}}{2a}\Rightarrow\frac{-1+7}{2\cdot1}=\frac{6}{2}=3[/tex]
