Odpowiedź :
[tex]2x(2x+1)=1\\4x^2+2x-1=0\\\\\Delta=2^2-4\cdot4\cdot(-1)=4+16=20\\\sqrt{\Delta}}=\sqrt{20}=2\sqrt5\\\\x_1=\dfrac{-2-2\sqrt5}{2\cdot4}=\dfrac{-1-\sqrt5}{4}\\\\x_2=\dfrac{-2+2\sqrt5}{2\cdot4}=\dfrac{-1+\sqrt5}{4}[/tex]
[tex]2x(2x+1)=1\\4x^2+2x-1=0\\\\\Delta=2^2-4\cdot4\cdot(-1)=4+16=20\\\sqrt{\Delta}}=\sqrt{20}=2\sqrt5\\\\x_1=\dfrac{-2-2\sqrt5}{2\cdot4}=\dfrac{-1-\sqrt5}{4}\\\\x_2=\dfrac{-2+2\sqrt5}{2\cdot4}=\dfrac{-1+\sqrt5}{4}[/tex]