[tex]Zadanie\\\\P=30\ cm^{2}\\a=?\\b=a+7\ cm=?\\c=?\\\\P=\frac{1}{2}\cdot a\cdot b\\\\ \frac{1}{2}\cdot a\cdot(a+7) =30\ \ \ \mid\cdot2\\\\a^{2}+7a=60\ \ \ \mid-60\\\\ a^{2}+7a-60=0\\\\ \Delta=b^{2}-4ac\\\\\Delta=49+240=289\\\\ \sqrt{\Delta} =\sqrt{289} = 17\\\\ x_1=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-7-17}{2\cdot1}=\frac{-24}{2}=-12\ -\ nie\ moze\ byc\\\\x_2=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-7+17}{2\cdot1}= \frac{10}{2}=5[/tex]
[tex]a=5\ cm\\\\b=a+7=5+7=12\ cm\\\\c^{2}=a^{2} +b^{2}\\\\c^{2}= 5^{2}+12^{2} \\\\c^{2}=25+ 144\\\\c^{2}=169\\\\c=\sqrt{169}\\\\c=13\ cm\\\\Odp.\ Dlugosci\ bokow\ trojkata\ wynosza:\ 5\ cm,12\ cm,13\ cm[/tex]
