Odpowiedź:
[tex]\huge\boxed{x\in\left\{-1,\ \dfrac{3-\sqrt{13}}{2},\ \dfrac{3+\sqrt{13}}{2}\right\}}[/tex]
Szczegółowe wyjaśnienie:
[tex]x^3-12x^2=-13\qquad|+13\\\\x^3-12x^2+13=0\\\\x^3+x^2-13x^2+13=0\\\\x^2(x+1)-13(x^2-1)=0\qquad|a^2-b^2=(a-b)(a+b)\\\\x^2(x+1)-13(x-1)(x+1)=0\\\\(x+1)\bigg[x^2-13(x-1)\bigg]=0\\\\(x+1)(x^2-13x+13)=0\iff x+1=0\ \vee\ x^2-13x+1=0\\\\\boxed{x=-1}\\\\a=1,\ b=-13,\ c=13\\\Delta=(-13)^2-4\cdot1\cdot13=169-52=117\\\sqrt\Delta=\sqrt{117}=\sqrt{9\cdot13}=\sqrt9\cdot\sqrt{13}=3\sqrt{13}\\\\x_1=\dfrac{-(-13)-3\sqrt{13}}{2\cdot1}=\dfrac{13-3\sqrt{13}}{2}\\\\x_2=\dfrac{-(-13)+3\sqrt{13}}{2\cdot1}=\dfrac{13+3\sqrt{13}}{2}[/tex]