Odpowiedź:
[tex](x+3)^2-5(x+3)+4=0\\\\x^2+6x+9-5x-15+4=0\\\\x^2+x-2=0\\\\\Delta=b^2-4ac\\\\\Delta=1^2-4*1*(-2)=1+8=9\\\\\sqrt{\Delta}=\sqrt{9}=3\\\\\\x_{1}=\dfrac{-b-\sqrt{\Delta} }{2a}=\dfrac{-1-3}{2*1}=\dfrac{-4}{2}=-2\\\\\\x_{2}=\dfrac{-b+\sqrt{\Delta} }{2a}=\dfrac{-1+3}{2*1}=\dfrac{2}{2}=1[/tex]
