Odpowiedź:
[tex]-x^2+8x-15=0\\\\\Delta=b^2-4ac=8^2-4\cdot(-1)\cdot(-15)=64-60=4\\\\\sqrt{\Delta}=\sqrt{4}=2\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-8-2}{2\cdot(-1)}=\frac{-10}{-2}=5\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-8+2}{2\cdot(-1)}=\frac{-6}{-2}=3[/tex]
[tex]12x^2-11x+2=0\\\\\Delta=b^2-4ac=(-11)^2-4\cdot12\cdot2=121-96=25\\\\\sqrt{\Delta}=\sqrt{25}=5\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2\cdot12}=\frac{11-5}{24}=\frac{6}{24}=\frac{1}{4}\\\\\\x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-(-11)+5}{2\cdot12}=\frac{11+5}{24}=\frac{16}{24}=\frac{2}{3}[/tex]
[tex]x^2+6x+5=0\\\\\Delta=b^2-4ac=6^2-4\cdot1\cdot5=36-20=16\\\\\sqrt{\Delta}=\sqrt{16}=4\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-6-4}{2\cdot1}=\frac{-10}{2}=-5\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-6+4}{2\cdot1}=\frac{-2}{2}=-1[/tex]
[tex]x^2-3x-4=0\\\\\Delta=b^2-4ac=(-3)^2-4\cdot1\cdot(-4)=9+16=25\\\\\sqrt{\Delta}=\sqrt{25}=5\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2\cdot1}=\frac{3-5}{2}=\frac{-2}{2}=-1\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2\cdot1}=\frac{3+5}{2}=\frac{8}{2}=4[/tex]
[tex]2x^2-3x=2\\\\2x^2-3x-2=0\\\\\Delta=b^2-4ac=(-3)^2-4\cdot2\cdot(-2)=9+16=25\\\\\sqrt{\Delta}=\sqrt{25}=5\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2\cdot2}=\frac{3-5}{4}=\frac{-2}{4}=-\frac{1}{2}\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2\cdot2}=\frac{3+5}{4}=\frac{8}{4}=2[/tex]
