Odpowiedź:
[tex](x-5)^2=25\\\\x^2-10x+25=25\\\\x^2-10x+25-25=0\\\\x^2-10x=0\\\\\Delta=b^2-4ac\\\\\Delta=(-10)^2-4\cdot1\cdot0=100-0=100\\\\\sqrt{\Delta}=\sqrt{100}=10\\\\\\x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10}{2\cdot1}=\frac{10-10}{2}=\frac{0}{2}=0\\\\x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10}{2\cdot1}=\frac{10+10}{2}=\frac{20}{2}=10\\\\x\in\lbrace{0,10\rbrace[/tex]
[tex]lub\\\\(x-5)^2=25\\\\(x-5)^2-25=0\\\\x^2-10x+25-25=0\\\\x^2-10x=0\\\\x(x-10)=0\\\\x=0\ \ \ \ \vee\ \ \ \ x-10=0\\\\x=0\ \ \ \ \vee\ \ \ \ x=10[/tex]
