Алгебра: (х²-3х)²-17(х²-3х)-18=0(через замену)

$(x^{2}-3x)^{2}-17(x^{2}-3x)-18=0\\ x^{2}-3x=t\\ t^{2}-17t-18=0\\ t_{1}+t_{2}=17;\space t_{1}t_{2}=-18\\ \left[\begin{array}{l} t=18\\ t=-1 \end{array}\right.\to \left[\begin{array}{l} x^{2}-3x=18\\ x^{2}-3x=-1 \end{array}\right. \to \left[\begin{array}{l} \left[\begin{array}{l} x=6\\ x=-3 \end{array}\right\\ \\ \left[\begin{array}{l} x=\frac{3+\sqrt{5}}{2}\\ x=\frac{3-\sqrt{5}}{2} \end{array}\right. \end{array}\right.\\$

Ответ: -3; 6; $\frac{3\pm \sqrt{5}}{2}$ .