Answer
The solutions are $x=3+\sqrt{19}$ and $x=3-\sqrt{19}$.
Work Step by Step
The given equation is
$\Rightarrow x^2-6x=10$
The coefficient of the $x^2-$term is $1$, and the coefficient of the $x-$term is an even number.
So, solve by completing the square.
$\Rightarrow x^2-6x=10$
Add $(\frac{-6}{2})^2=(-3)^2=9$ to each side.
$\Rightarrow x^2-6x+9=10+9$
Write the left side as the square of a binomial.
$\Rightarrow (x-3)^2=19$
Take the square root of each side.
$\Rightarrow x-3=\pm\sqrt{19}$
Add $3$ to each side.
$\Rightarrow x-3+3=3\pm\sqrt{19}$
Simplify.
$\Rightarrow x=3\pm\sqrt{19}$
The solutions are $x=3+\sqrt{19}$ and $x=3-\sqrt{19}$.
