Answer
The solution set is $\left\{3-\sqrt{22}, 3+\sqrt{22}\right\}$.
Work Step by Step
RECALL:
A quadratic equation in the form $x^2 + bx =c$ can be solved by completing the square by adding $\left(\frac{b}{2}\right)^2$ to both sides of the equation then using the square root property.
The equation will become $(x+\frac{b}{2})^2=c+(\frac{b}{2})^2$
Complete the square by adding $(\frac{-6}{2})^2=9$ to both sides of the equation to obtain:
$x^2-6x+9=13+9
\\x^2-6x+9=22$
Write the trinomial as a square of a binomial to obtain:
$(x-3)^2=22$
Take the square root of both sides to obtain:
$x-3 = \pm\sqrt{22}$
Add $3$ to both sides to obtain:
$x-3+3 = 3 \pm \sqrt{22}
\\x = 3\pm \sqrt{22}$
Split the solutions to obtain:
$x_1=3+\sqrt{22}$
$\\x_2=3-\sqrt{22}$
Thus, the solution set is $\left\{3-\sqrt{22}, 3+\sqrt{22}\right\}$.
