Answer:

[tex]\boxed{x = \frac{1}{3} \pm \frac{2 \sqrt{3} }{3}}[/tex]

Step-by-step explanation:

All equations of the form ax² + bx + c = 0 can be solved using the quadratic formula:

[tex]\boxed{ \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} } \\ [/tex]

Given:

• [tex] {9x}^{2} - 11 = 6x[/tex]

We to make one side equal to 0 Subract 6x from both sides to get 9x² - 6x - 11 = 0.

Therefore,

a = 9

b = -6

c = -11

Now, substitute it to the formula. Therefore,

[tex]x = \frac{ - ( - 6) \pm \sqrt{( -6 ) {}^{2} - 4(9)(11) } }{2(9)} \\ [/tex]

[tex]x = \frac{6 \pm \sqrt{36 + 396} }{18} \\ [/tex]

[tex]x = \frac{6 \pm \sqrt{432} }{18} \\ [/tex]

[tex]x = \frac{6 \pm12 \sqrt{3} }{18} \\ [/tex]

[tex]x = \frac{1 \pm2 \sqrt{3} }{3} \\ [/tex]

[tex]x = \frac{1}{3} \pm \frac{2 \sqrt{3} }{3} \\ [/tex]

Hence,

[tex]\boxed{x = \frac{1}{3} + \frac{2 \sqrt{3} }{3} , \: \: x = \frac{1}{3} - \frac{2 \sqrt{3} }{3}} [/tex]

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