Solve the equatinn \( 6 x^{4}+8 x^{2}=26 x^{3} \)

To solve the equation \( 6x^{4} + 8x^{2} = 26x^{3} \), first, let's rearrange it into standard polynomial form: \[ 6x^{4} - 26x^{3} + 8x^{2} = 0. \] Now, we can factor out the common term \( 2x^{2} \): \[ 2x^{2}(3x^{2} - 13x + 4) = 0. \] This gives us two factors: 1. \( 2x^{2} = 0 \) which yields \( x = 0 \). 2. For the quadratic \( 3x^{2} - 13x + 4 = 0 \), we can solve it using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}, \] where \( a = 3, b = -13, c = 4 \). Plugging in these values: \[ x = \frac{13 \pm \sqrt{(-13)^{2} - 4 \cdot 3 \cdot 4}}{2 \cdot 3} = \frac{13 \pm \sqrt{169 - 48}}{6} = \frac{13 \pm \sqrt{121}}{6} = \frac{13 \pm 11}{6}. \] This leads to two potential solutions: \[ x = \frac{24}{6} = 4 \quad \text{and} \quad x = \frac{2}{6} = \frac{1}{3}. \] Thus, the complete set of solutions is: \[ x = 0, \quad x = 4, \quad x = \frac{1}{3}. \]