Solve the following quadratic equation using the quadratic formula. \[ -6 x^{2}+3=4 x \]

First, let's rearrange the equation into standard form, \(ax^2 + bx + c = 0\). Start by moving all terms to one side: \[-6x^2 - 4x + 3 = 0.\] Now, we can identify the coefficients: \(a = -6\), \(b = -4\), and \(c = 3\). Plug these values into the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Calculating the discriminant: \[b^2 - 4ac = (-4)^2 - 4(-6)(3) = 16 + 72 = 88.\] Now substitute back into the quadratic formula: \[x = \frac{-(-4) \pm \sqrt{88}}{2(-6)} = \frac{4 \pm \sqrt{88}}{-12}.\] We can simplify \(\sqrt{88}\) to \(2\sqrt{22}\). Therefore, we have: \[x = \frac{4 \pm 2\sqrt{22}}{-12} = \frac{4 \pm 2\sqrt{22}}{-12} = \frac{-2 \mp \sqrt{22}}{6}.\] Thus, the solutions are: \[x = \frac{-2 + \sqrt{22}}{6}, \quad x = \frac{-2 - \sqrt{22}}{6}.\]