3х2 - 7х + 4 = 0

To solve the quadratic equation \(3x^2 - 7x + 4 = 0\), we can use the quadratic formula, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 3\), \(b = -7\), and \(c = 4\). First, we calculate the discriminant (\(D\)): \[ D = b^2 - 4ac \] \[ D = (-7)^2 - 4 \cdot 3 \cdot 4 \] \[ D = 49 - 48 \] \[ D = 1 \] Since the discriminant is positive, there will be two distinct real roots. Now we can substitute \(a\), \(b\), and \(D\) into the quadratic formula: \[ x = \frac{-(-7) \pm \sqrt{1}}{2 \cdot 3} \] \[ x = \frac{7 \pm 1}{6} \] This gives us two solutions: 1. For the plus case: \[ x_1 = \frac{7 + 1}{6} = \frac{8}{6} = \frac{4}{3} \] 2. For the minus case: \[ x_2 = \frac{7 - 1}{6} = \frac{6}{6} = 1 \] Thus, the solutions to the equation \(3x^2 - 7x + 4 = 0\) are: \[ x_1 = \frac{4}{3}, \quad x_2 = 1 \]