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[tex]0,3x^2+0,7x-0,1=0\ /\cdot 10\\\\3x^2+7x-1=0\\\\a= 3\\b=7\\c=-1\\\\1.\ Licze\ delte:\\\Delta =b^2-4ac=7^2-4\cdot 3\cdot (-1)=49+12=61>0\ dwa\ rozwiazania\\\\2.\ Obliczam\ pierwiastki\ rownania:\\x_1=\frac{-b-\sqrt{\Delta} }{2a} =\frac{-7-\sqrt{61} }{6} \\x_2=\frac{-b+\sqrt{\Delta} }{2a} =\frac{-7+\sqrt{61} }{6}\\\\3.\ Obliczam\ sume\ kwadratow\ pierwiastkow\ rownania:\\x_1^2+x_2^2=(\frac{-7-\sqrt{61} }{6})^2+( \frac{-7+\sqrt{61} }{6})^2=[/tex]
[tex]=\frac{49+14\sqrt{61}+61}{36}+\frac{49-14\sqrt{61}+61}{36}=\\=\frac{49+14\sqrt{61}+61+49-14\sqrt{61}+61}{36} =\\=\frac{220}{36}=\frac{55}{9}=\boxed{6\frac{1}{9}}[/tex]
ODP.: Suma kwadratów pierwiastków tego równania wynosi [tex]6\frac{1}{9}[/tex]