9x²−12x−14=0 solve by completing the square

$$\frac{9x^2}{9} - \frac{12x}{9} - \frac{14}{9} = \frac{0}{9}$$ $$x^2 - \frac{4}{3}x - \frac{14}{9} = 0$$

Take half of the coefficient of $x$ (which is $-\frac{4}{3}$), square it, and add it to both sides.
Half of $-\frac{4}{3}$ is $-\frac{4}{3} \times \frac{1}{2} = -\frac{2}{3}$.
Squaring it gives $\left(-\frac{2}{3}\right)^2 = \frac{4}{9}$.

$$x^2 - \frac{4}{3}x + \frac{4}{9} = \frac{14}{9} + \frac{4}{9}$$

$$\left(x - \frac{2}{3}\right)^2 = \frac{18}{9}$$ $$\left(x - \frac{2}{3}\right)^2 = 2$$

$$x = \frac{2}{3} \pm \sqrt{2}$$ $$x = \frac{2 \pm 3\sqrt{2}}{3}$$

The solutions are:

$$x_1 = \frac{2 + 3\sqrt{2}}{3}$$ $$x_2 = \frac{2 - 3\sqrt{2}}{3}$$