√3x^2-6x+√3=0 Please solve.

I suck at working with square roots in calculations and need more practice, but would really like to see this solved.

Have you tried plugging it into the quadratic formula?
I tried factorising, looking for a way to get rid of the square root but it just ended up being that the quadratic formula was way better to use.
Plug it in and this is what you get:

[--6 +- sqrt(36 - 4 (sqrt3 x sqrt3) ] / 2sqrt3
Simplify things a little. The sqrt3 x sqrt3 will become JUST 3
[6 +- sqrt(36 - 4(3)] / 2sqrt3
[6+- sqrt(24)] / 2sqrt3
Now simplify the sqrt24. You can make that sqrt4 x sqrt6. sqrt4 is just 2. So you get 2sqrt6
(6 +- 2sqrt6) / 2sqrt3
Now let's rationalise by multiplying top and bottom by sqrt3. We get
6sqrt3 +- 2sqrt18 / 2sqrt3xsqrt3
Simplify the bottom of the fraction. We get:
6sqrt3 +- 2sqrt18 / 2 x 3
6sqrt3 +- 2sqrt18 / 6
Now simplify the sqrt 18. We can make that sqrt9 x sqrt2. sqrt9 is just 3. So it becomes:
6sqrt3 +- 2 x 3 x sqrt2 / 6
Simplify
6sqrt3 +- 6sqrt2 / 6
Now we can cancel out the 6's on the top and bottom to simply leave us with:
sqrt 3+- sqrt2


I hope that helps.
Jamz159

(6±√36-4(√3)(√3))/2√3=(6±√24)/2√3=(6±2√6…‡ You can then eliminate the 2 in the denominator in all terms. You're left with (3±√6)/√3. Rationalizing the denominator yields (3√3±3√2)/3, or just √3±√2.

I am assuming that the expression is:

√(3x² - 6x) + √(3) = 0

Since both radicands are both positive, you can't eliminate both of them to get 0. Therefore, there are no solutions.

I hope this helps!