Odpowiedź:
[tex]f(x)=2x^2+5x-3\\\\\\a)\\\\2x^2+5x-3=0\\\\a=2\ \ ,\ \ b=5\ \ ,\ \ c=-3\\\\\Delta=b^2-4ac=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}{4}=\frac{2}{4}=\frac{1}{2}[/tex]
[tex]b)\\\\p=-\frac{b}{2a}=-\frac{5}{2\cdot2}=-\frac{5}{4}\\\\q=-\frac{\Delta}{4a}=-\frac{49}{4\cdot2}=-\frac{49}{8}\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=2(x-(-\frac{5}{4}))^2+(-\frac{49}{8})\\\\f(x)=2(x+\frac{5}{4})^2-\frac{49}{8}[/tex]
