## Trigonometry (11th Edition) Clone

The solution set of this problem, written in standard form, is $$\Big\{\frac{1}{3}\pm\frac{\sqrt6}{3}i\Big\}$$
$$3(3x^2-2x)=-7$$ The equation is not in standard form, so first, we need to bring it back to standard form: $$9x^2-6x=-7$$ $$9x^2-6x+7=0$$ Now the equation is in standard form, the quadratic formula can be applied. $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ As $a=9, b=-6, c=7$ $$x=\frac{-(-6)\pm\sqrt{(-6)^2-4\times9\times7}}{2\times9}$$ $$x=\frac{6\pm\sqrt{36-252}}{18}$$ $$x=\frac{6\pm\sqrt{-216}}{18}$$ Now we rewrite $\sqrt{-216}=i\sqrt{216}=i\sqrt{36\times6}=6i\sqrt6$ $$x=\frac{6\pm 6i\sqrt6}{18}$$ Finally, we simplify $$x=\frac{1\pm i\sqrt6}{3}$$ The solution set of this problem, written in standard form, is $$\Big\{\frac{1}{3}\pm\frac{\sqrt6}{3}i\Big\}$$