## Precalculus (6th Edition)

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