We are asked to solve the quadratic equation $3x^2 - 5x + 1 = 0$ for $x \in \mathbb{R}$, and we must give the answer correct to 2 decimal places.

1. Problem Description

We are asked to solve the quadratic equation $3x^2 - 5x + 1 = 0$ for $x \in \mathbb{R}$, and we must give the answer correct to 2 decimal places.

2. Solution Steps

We can use the quadratic formula to solve the equation $ax^2 + bx + c = 0$:

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

In our case, $a = 3$, $b = -5$, and $c = 1$. Plugging these values into the quadratic formula, we get:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(1)}}{2(3)}$

$x = \frac{5 \pm \sqrt{25 - 12}}{6}$

$x = \frac{5 \pm \sqrt{13}}{6}$

So, we have two solutions:

$x_1 = \frac{5 + \sqrt{13}}{6}$

$x_2 = \frac{5 - \sqrt{13}}{6}$

We need to approximate these values to 2 decimal places.

$\sqrt{13} \approx 3.60555$

$x_1 = \frac{5 + 3.60555}{6} = \frac{8.60555}{6} \approx 1.43426 \approx 1.43$

$x_2 = \frac{5 - 3.60555}{6} = \frac{1.39445}{6} \approx 0.232408 \approx 0.23$

3. Final Answer

The solutions are $x \approx 1.43$ and $x \approx 0.23$.