Please, help me to solve this equation. No advanced math should be needed.

$$3x^2 - 4x + \sqrt{3x^2 - 4x - 6} = 18$$

I'm clueless. It should be simple.

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What have you tried so far? Certainly the square root sign is a nuisance. Can you try to get rid of it? –  Dylon Chow Jul 24 '12 at 20:55
To simplify, let $z=3x^2-4x-6$. Then your equation is $z+\sqrt z=12$. –  David Mitra Jul 24 '12 at 20:56
Denote by $y=\sqrt{3x^2-x-6}$ –  Tigran Hakobyan Jul 24 '12 at 20:57
@DavidMitra Wow, that's super smart! I would've never thought of that! I'll try it! –  BeetleTheNeato Jul 24 '12 at 20:59
@BeetleTheNeato Several answers provided below are smarter :) –  David Mitra Jul 24 '12 at 21:00
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Hint : set $u:= \sqrt{3 x^2-4x-6}$ then your equation becomes $u^2+6+u=18$ or $$u^2+u-12=0$$

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• Denote $u = \sqrt{3 x^2 - 4x -6}$.
• Write the original equation in terms of $u$. You should get a quadratic equation.
• Solve it.
• For each root $u_\ast$ found, solve $3 x^2 - 4x -6 = u_\ast^2$.
• Verify your solutions against the original equation and weed out extraneous solutions.
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Put $u = 3x^2 - 4x$. Then the equation becomes $$u + \sqrt{u - 6} = 18.$$ Once you solve for $u$ in the auxillary equation, you are left with a quadratic.