Срочно нужно! Распишите как вы решали

$\left \{ {{x^2+y^2=17} \atop {5x-3y=17}} \right. \left \{ {{\frac{9y^2+102y+289}{25}+y^2=17} \atop {x=\frac{3y+17}{5}}} \right. \left \{ {{9y^2+102y+289+25y^2=425} \atop {x=0,6y+3,4}} \right. \\34y^2+102y-136=0\\y^2+3y-4=0\\D=9+16=25=5^2\\y_{1,2}=\frac{-3б5}{2} =\left | {{1} \atop {-4}} \right. \\ \left \{ {{y_1=1} \atop {x_1=0,6+3,4}} \right. \left \{ {{y_1=1} \atop {x_1=4}} \right. \left \{ {{y_2=-4} \atop {x_2=-2,4+3,4}} \right. \left \{ {{y_2=-4} \atop {x_2=1}} \right.$

Ответ: $(4;1);(1;-4)$