Odpowiedź:
[tex]a)\\\\x^2-49=0\\\\(x-7)(x+7)=0\\\\x-7=0\ \ \ \ \vee\ \ \ \ x+7=0\\\\x=7\ \ \ \ \ \ \ \ \vee\ \ \ \ x=-7\\\\\\b)\\\\x^2+7x=0\\\\x(x+7)=0\\\\x=0\ \ \ \ \vee\ \ \ \ x+7=0\\\\x=0\ \ \ \ \vee\ \ \ \ x=-7[/tex]
[tex]c)\\\\x^2+17x+16=0\\\\\Delta=b^2-4ac\\\\\Delta=17^2-4\cdot1\cdot16=289-64=225\\\\\sqrt{\Delta}=\sqrt{225}=15\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-17-15}{2\cdot1}=\frac{-32}{2}=-16\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-17+15}{2\cdot1}=\frac{-2}{2}=-1[/tex]
[tex]d)\\\\2x^2+5x-3=0\\\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot2\cdot(-3)=25+24=49\\\\\sqrt{\Delta}=\sqrt{49}=7\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-5-7}{2\cdot2}=\frac{-12}{4}=-3\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-5+7}{2\cdot2}=\frac{2}{4}=\frac{1}{2}[/tex]
[tex]e)\\\\3x^2+2x+1=0\\\\\Delta=b^2-4ac\\\\\Delta=2^2-4\cdot3\cdot1=4-12=-8\\\\\Delta<0 \ \ brak\ \ rozwiazania[/tex]
[tex]f)\\\\2x^2+17x+1=0\\\\\Delta=b^2-4ac\\\\\Delta=17^2-4\cdot2\cdot1=289-8=281\\\\\sqrt{\Delta}=\sqrt{281}\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-17-\sqrt{281}}{2\cdot2}=\frac{-17-\sqrt{281}}{4}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-17+\sqrt{281}}{2\cdot2}=\frac{-17+\sqrt{281}}{4}[/tex]
