Question

$2 {x}^{2} - 4x = 1 + \sqrt{13}$
$2 {x}^{2} - 4x - (1 + \sqrt{13} ) = 1$ ​

Answer 1

$1)..2x^2-4x=1+\sqrt{13}\\2x^2-4x-1-\sqrt{13}=0\\x=\frac{-(-4)\pm\sqrt{(-4)^2-4\cdot2(-1-\sqrt{13})}}{2\cdot2} \\x=\frac{4\pm\sqrt{16-8(-1-\sqrt{13})}}{4}\\x=\frac{4\pm\sqrt{16+8+8\sqrt{13}}}{4}\\x=\frac{4\pm\sqrt{24+8\sqrt{13}}}{4}\\x=\frac{4\pm2\sqrt{6+2\sqrt{13}}}{4} \\ x=\frac{4+2\sqrt{6+2\sqrt{13} } }{4}\\x=\frac{4-2\sqrt{6+2\sqrt{13} } }{4}\\x=\frac{2+\sqrt{6+2\sqrt{13} } }{2}\\x=\frac{2-\sqrt{6+2\sqrt{13} } }{2}\\x_1=\frac{2-\sqrt{6+2\sqrt{13} } }{2},x_2=x_1=\frac{2+\sqrt{6+2\sqrt{13} } }{2}$