Answer
$x=1+\sqrt{7}$
Work Step by Step
Set up the equation:
Let $x$ be the number. Note that $x>0$.
$x^2 - (6+2x)=0$
$x^2 - 6 - 2x =0$
$x^2 -2x-6=0$
$x^2 -2x=6$
Compute for $x$ by completing the square.
The coefficient of the $x$-term is $-2$; $\frac{-2}{2}=-1$; $-1^2=1$
Add $1$ to both sides to complete the square.
$x^2 -2x+1=6+1$
$x^2 -2x+1=7$
$(x-1)^2=7$
$x-1=±\sqrt{7}$
$x=1±\sqrt{7}$
Since $x>0$, thus, $x=1+\sqrt{7}$.
