Answer
The answer is $1+\sqrt{7}$
Work Step by Step
When the sum of $6$ and twice a positive number is subtracted from 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:
$x^{2}-(2x+6)=0$
Simplify the equation:
$x^{2}-2x-6=0$
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 $c=-6$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-(-2)\pm\sqrt{(-2)^{2}-4(1)(-6)}}{2(1)}=\dfrac{2\pm\sqrt{4+24}}{2}=...$
$...=\dfrac{2\pm\sqrt{28}}{2}=\dfrac{2\pm2\sqrt{7}}{2}=1\pm\sqrt{7}$
Since the number to be found is positive. The answer is $1+\sqrt{7}$
