Answer
The answer is $1-\sqrt{2}$
Work Step by Step
When the sum of $1$ and twice a negative number is subtracted from twice the square of the number, $0$ results. Find the number.
Let $x$ be the number to be found
Write an equation that represents the problem:
$x^{2}-(2x+1)=0$
Simplify the equation:
$x^{2}-2x-1=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. In this case, $a=1$, $b=-2$ and $c=-1$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-1)}}{2(1)}=\dfrac{2\pm\sqrt{4+4}}{2}=...$
$...=\dfrac{2\pm\sqrt{8}}{2}=\dfrac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2}$
Since the number to be found is negative, the answer is $1-\sqrt{2}$
