Найдите x в квадрате + y в квадрате если x+y = 7 и x y =3

Решите задачу:

$\left \{ {{x+y=7} \atop {xy=3}} \right. =\ \textgreater \ \left \{ {{x=7-y} \atop {(7-y)*y=3}} \right. \\(7-y)y=3\\7y-y^2=3\\y^2-7y+3=0\\D=(-7)^2-4*1*3=49-12=37\\y_1= \frac{7+ \sqrt{37} }{2};\; \; \; \; y_2= \frac{7- \sqrt{37} }{2}\\x_1= 7-y_1=7-\frac{7+ \sqrt{37} }{2}= \frac{14-7- \sqrt{37} }{2}= \frac{7- \sqrt{37} }{2}\\\\x_2= 7-y_2=7-\frac{7- \sqrt{37} }{2}= \frac{14-7+ \sqrt{37} }{2}= \frac{7+ \sqrt{37} }{2} \\\\x_1^2+y_1^2= (\frac{7- \sqrt{37} }{2})^2+(\frac{7+ \sqrt{37} }{2})^2= \frac{49+37-14 \sqrt{37}+49+37+14 \sqrt{37} }{4}= \frac{172}{4}=43\\\\x_2^2+y_2^2= (\frac{7+ \sqrt{37} }{2})^2+(\frac{7- \sqrt{37} }{2})^2= \frac{49+37+14 \sqrt{37}+49+37-14 \sqrt{37} }{4}= \frac{172}{4}=43\\\\x^2+y^2=43$