Odpowiedź:
[tex]a)\ \ x^2-3x-18\geq 0\\\\\Delta=b^2-4ac=(-3)^2-4*1*(-18)=9+72=81\\\\\sqrt{\Delta}=\sqrt{81}=9\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-9}{2*1}=\frac{3-9}{2}=\frac{-6}{2}=-3\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+9}{2*1}=\frac{3+9}{2}=\frac{12}{2}=6[/tex]
[tex]b)\\\\-x^2-5x>0\\\\\Delta=b^2-4ac=(-5)^2-4*(-1)*0=25+0=25\\\\\sqrt{\Delta}=\sqrt{25}=5\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*(-1)}=\frac{5-5}{-2}=\frac{0}{-2}=0\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*(-1)}=\frac{5+5}{-2}=\frac{10}{-2}=-5[/tex]
[tex]c)\\\\(4-2x)^2=16\\\\16-16x+4x^2=16\\\\16-16x+4x^2-16=0\\\\4x^2-16x=0\\\\\Delta=b^2-4ac=(-16)^2-4*4*0=256-0=256\\\\\sqrt{\Delta}=\sqrt{256} =16\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*4}=\frac{16-16}{8}=\frac{0}{8}=0\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*4}=\frac{16+16}{8}=\frac{32}{8}=4[/tex]
