#### Answer

$x=\left\{-3+\frac{\sqrt2}{2},-3-\frac{\sqrt2}{2}\right\}$

#### Work Step by Step

Distribute $x$ to obtain:
$x(x)+x(6)=9
\\x^2+6x=9$
Subtract $9$ to both sides:
$x^2+6x-9=9-9
\\x^2+6x-9=0$
The equation above has $a=1, b=6$ and $c=-9$.
Solve using the quadratic formula to obtain:
$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}
\\x=\dfrac{-6 \pm \sqrt{6^2-4(1)(-9)}}{2(1)}
\\x=\dfrac{-6 \pm \sqrt{36+36}}{2}
\\x=\dfrac{-6 \pm \sqrt{72}}{2}
\\x=\dfrac{-6\pm\sqrt{36(2)}}{2}
\\x=\dfrac{-6\pm 6\sqrt2}{2}
\\x=\dfrac{-6}{2} \pm \dfrac{\sqrt{2}}{2}
\\x=-3 \pm \dfrac{\sqrt2}{2}$
Thus, the solutions are:
$x=-3+\frac{\sqrt2}{2}$ nd $x=-3-\frac{\sqrt2}{2}$