Разность квадратов корней уравнения Х2 -17Х +q=0 равно 85.найдите q

[latex]x^2-17x+q=0\\\\ \left \{ {{x_1^2-x_2^2=85} \atop {x_1+x_2=17\; (teor.\; Vieta)}} \right. \; \left \{ {{(x_1-x_2)(x_1+x_2)=85} \atop {x_1+x_2=17}} \right. \; \left \{ {{17(x_1-x_2)=85} \atop {x_1+x_2=17}} \right. \\\\ \left \{ {{x_1-x_2=\frac{85}{7}} \atop {x_1+x_2=17}} \right. \; \left \{ {{2x_1=\frac{85}{17}+17=\frac{374}{17}} \atop {2x_2=17-\frac{85}{17}=\frac{204}{17}}} \right. \; \left \{ {{x_1=\frac{187}{17}} \atop {x_2=\frac{102}{17}}} \right. \\\\q=x_1\cdot x_2[/latex] [latex]q= \frac{187}{17} \cdot \frac{102}{17} = \frac{19074}{289} =66\\\\x^2-17x+66=0[/latex]