Odpowiedź:
[tex]x(x-3)=3x-4\\\\x^2-3x=3x-4\\\\x^2-3x-3x+4=0\\\\x^2-6x+4=0\\\\\Delta=b^2-4ac=(-6)^2-4*1*4=36-16=20\\\\\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=2\sqrt{5}\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{5}}{2*1}=\frac{6-2\sqrt{5}}{2}=\frac{\not2(3-\sqrt{5})}{\not2}=3-\sqrt{5}\\\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{5}}{2*1}=\frac{6+2\sqrt{5}}{2}=\frac{\not2(3+\sqrt{5})}{\not2}=3+\sqrt{5}[/tex]