Odpowiedź:
[tex]2x^2-5x-3=0\\\\a=2\ \ ,\ \ b=-5\ \ ,\ \ c=-3\\\\\\\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{5-7}{4}=\frac{-2}{4}=-\frac{1}{2}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2\cdot2}=\frac{5+7}{4}=\frac{12}{4}=3[/tex]
[tex]R\'ownanie\ \ ma\ \ dwa\ \ rozwiazania\ \ x_{1}=-\frac{1}{2}\ \ ,\ \ x_{2}=3[/tex]
