Roots of the quadratic equation

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).    

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