Okay, here we have this:
Considering the provided information, we obtain the following equation:
[tex]\begin{gathered} 2x^2-\mleft(1+2x\mright)=0 \\ 2x^2-1-2x=0 \end{gathered}[/tex]Let's solve it to find the number:
[tex]\begin{gathered} x_{1,\: 2}=\frac{-\left(-2\right)\pm\sqrt{\left(-2\right)^2-4\cdot\:2\left(-1\right)}}{2\cdot\:2} \\ x_{1,\: 2}=\frac{-\left(-2\right)\pm\:2\sqrt{3}}{2\cdot\:2} \\ x_1=\frac{-\left(-2\right)+2\sqrt{3}}{2\cdot\:2},\: x_2=\frac{-\left(-2\right)-2\sqrt{3}}{2\cdot\:2} \\ x=\frac{1+\sqrt{3}}{2},\: x=\frac{1-\sqrt{3}}{2} \end{gathered}[/tex]Since the number is negative, then we are left with the second result, so finally we find that the number is:
[tex]x=\frac{1-\sqrt[]{3}}{2}[/tex]