Answer
The solution set is\[\left\{ -1\pm \sqrt{3} \right\}\].
Work Step by Step
Consider the required number as\[x\], then, according to the given condition, \[2{{x}^{2}}-\left( 1+2\left( -x \right) \right)=0\].
The given equation can be written as
\[2{{x}^{2}}-1+2x=0\]
or
\[2{{x}^{2}}+2x-1=0\]
Compare the given equation with the equation\[a{{x}^{2}}+bx+c=0\].
Here,\[a=2,\ b=2,\text{ and }c=-1\].
Now, put these values in the quadratic formula\[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].
That is,
\[\begin{align}
& x=\frac{-2\pm \sqrt{{{\left( 2 \right)}^{2}}-4\times 2\times \left( -1 \right)}}{2\times 2} \\
& =\frac{-2\pm \sqrt{4+8}}{4} \\
& =\frac{-2\pm \sqrt{12}}{4} \\
& =\frac{-2\pm 2\sqrt{3}}{4}
\end{align}\]
Further simplify it to get
\[x=\frac{-1\pm \sqrt{3}}{2}\]
