3х*2-х=1-3х*2+х

Let's solve the equation step by step:

Given equation: \[ 3x^2 - x = 1 - 3x^2 + x \]

First, let's move all terms to one side of the equation:

\[ 3x^2 - x + 3x^2 - x - 1 = 0 \]

Combining like terms, we get:

\[ 6x^2 - 2x - 1 = 0 \]

Now, we can apply the quadratic formula, which is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our equation \( a = 6 \), \( b = -2 \), and \( c = -1 \).

Now plug in the values:

1. Calculate the discriminant \( b^2 - 4ac \):

\[ (-2)^2 - 4 \cdot 6 \cdot (-1) = 4 + 24 = 28 \]

2. Now substitute into the quadratic formula:

\[ x = \frac{-(-2) \pm \sqrt{28}}{2 \cdot 6} \]

\[ x = \frac{2 \pm \sqrt{28}}{12} \]

3. Simplify \( \sqrt{28} = 2\sqrt{7} \):

\[ x = \frac{2 \pm 2\sqrt{7}}{12} \]

\[ x = \frac{1 \pm \sqrt{7}}{6} \]

Thus, the solutions to the equation are:

\[ x = \frac{1 + \sqrt{7}}{6} \quad \text{and} \quad x = \frac{1 - \sqrt{7}}{6} \]