**x ^2 / 3 + 3 = 2x**

First we will multiply by 3:

==> 3x^2 / 3 + 3*3 = 3*2x

==> x^2 + 9 = 6x

Now we wqill subtract 6x from both sides:

==> x^2 - 6x + 9 = 0

Now we have two options to solve:

option (1) : factoring:

==> ( x62 - 6x + 9) = ( x-3)^2

==> (x-3)^2 = 0

==> (x-3) = 0

**==> x = 3**

Option (2): use the formula:

x = [ -b +- sqrt(b^2 - 4ac)]/2a

= [ 6 + sqrt(39 - 36) / 2 = 6/2

**==> x = 3**

We have to solve x^2/ 3 + 3 = 2x

First multiply all the terms with 3

=> x^2 + 9 = 6x

subtract 6x from both the sides

=> x^2 - 6x + 9 =0

We know that (a-b)^2 = a^2 + b^2 -2*a*b.

If we look at x^2 - 6x + 9 =0, we see that it can be written as (x - 3)^2 =0

Therefore x - 3 = 0

Or x = 3

**Therefore the required solution of x is 3. The quadratic equation has two equal roots.**

x^2/3 +3 = 2x.

To find all the solutions.

We multiply the given equation by 3.

x^2+3*3 = 3*2x = 6x.

x^2+9 = 6x.

x^2-6x+9 = 0.

(x-3)^2 = 0.

x-3 = 0

So x= 3.

Therefore x = 3 is the solution.

First, we'll eliminate the denominator by multiplying all terms in the equation by 3.

x^2*3/ 3 + 3*3 = 2*3*x

We'll simplify and we'll get:

x^2 + 9 = 6x

We'll subtract 6x both sides so we'll re-write the equation with the right term equal to zero. We'll re-arrange the terms:

x^2 - 6x + 9 = 0

We recognize the form of the binomial (x-3)^2:

(x-3)^2 = 0

x1 = x2 = 3

**The equation will have 2 equal real solutions: x1 = x2 = 3. **

Verify:

3^2/3 + 3 = 2*3

9/3 + 3 = 6

3 + 3 = 6

6 = 6