Odpowiedź:
[tex]a)\\\\-8x^2+5-\frac{1}{2}=\frac{1}{2}-x\\\\-8x^2+4\frac{1}{2}-\frac{1}{2}+x=0\\\\-8x^2+4+x=0\\\\-8x^2+x+4=0\ \ /\cdot(-1)\\\\8x^2-x-4=0\\\\a=8\ \ ,\ \ b=-1\ \ ,\ \ c=-4\\\\\Delta=b^2-4ac\\\\\Delta=(-1)^2-4\cdot8\cdot(-4)=1+128=129\\\\\sqrt{\Delta}=\sqrt{129}\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{129}}{2\cdot8}=\frac{1-\sqrt{129}}{16}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{129}}{2\cdot8}=\frac{1+\sqrt{129}}{16}[/tex]
[tex]b)\\\\-19x^2-10x+2=x^2+x-1\\\\-19x^2-10x+2-x^2-x+1=0\\\\-20x^2-11x+3=0\ \ /\cdot(-1)\\\\20x^2+11x-3=0\\\\a=20\ \ ,\ \ b=11\ \ ,\ \ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=11^2-4\cdot20\cdot(-3)=121+240=\sqrt{361}\\\\\sqrt{\Delta}=\sqrt{361}=19\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-11-19}{2\cdot20}=\frac{-30}{40}=-\frac{3}{4}\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-11+19}{2\cdot20}=\frac{8}{40}=\frac{1}{5}[/tex]
