To find the roots of the quadratic equation (3x^2 + 4x + 1 = 0), we can use the quadratic formula:

The quadratic formula is given by: (x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a})

In this equation, (a = 3), (b = 4), and (c = 1). Plugging these values into the formula:

(x = \frac{-4 \pm \sqrt{4^2 – 431}}{2*3})

(x = \frac{-4 \pm \sqrt{16 – 12}}{6})

(x = \frac{-4 \pm \sqrt{4}}{6})

(x = \frac{-4 \pm 2}{6})

Now, we have two possible solutions for x:

1. When using the plus sign: (x_1 = \frac{-4 + 2}{6} = \frac{-2}{6} = -\frac{1}{3})

2. When using the minus sign: (x_2 = \frac{-4 – 2}{6} = \frac{-6}{6} = -1)

Therefore, the roots of the quadratic equation (3x^2 + 4x + 1 = 0) are (x = -\frac{1}{3}) and (x = -1).