Use the quadratic formula to solve for x. 6x^2-8x+1=0 Round your answer to the nearest hundredth. If there is more than one solution, separate them with commas. x= square

To solve the quadratic equation $6x^{2}-8x+1=0$ using the quadratic formula, we can use the formula:<br /><br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a$, $b$, and $c$ are the coefficients of the quadratic equation.<br /><br />In this case, $a = 6$, $b = -8$, and $c = 1$.<br /><br />Plugging these values into the formula, we get:<br /><br />$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(6)(1)}}{2(6)}$<br /><br />Simplifying further:<br /><br />$x = \frac{8 \pm \sqrt{64 - 24}}{12}$<br /><br />$x = \frac{8 \pm \sqrt{40}}{12}$<br /><br />$x = \frac{8 \pm 2\sqrt{10}}{12}$<br /><br />$x = \frac{4 \pm \sqrt{10}}{6}$<br /><br />Therefore, the solutions to the quadratic equation $6x^{2}-8x+1=0$ are:<br /><br />$x = \frac{4 + \sqrt{10}}{6}$ and $x = \frac{4 - \sqrt{10}}{6}$<br /><br />Rounding these solutions to the nearest hundredth, we get:<br /><br />$x = 0.54$ and $x = 0.16$<br /><br />So, the final answer is:<br /><br />$x = 0.54, 0.16$