3x²-7x+4=0

To solve the quadratic equation \(3x^2 - 7x + 4 = 0\), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a\), \(b\), and \(c\) are the coefficients from the equation \(ax^2 + bx + c = 0\). In this case: - \(a = 3\) - \(b = -7\) - \(c = 4\) Now, we will calculate the discriminant (\(b^2 - 4ac\)): \[ b^2 - 4ac = (-7)^2 - 4 \cdot 3 \cdot 4 = 49 - 48 = 1 \] Since the discriminant is positive, we will have two distinct real roots. Now, applying the quadratic formula: \[ x = \frac{-(-7) \pm \sqrt{1}}{2 \cdot 3} \] \[ x = \frac{7 \pm 1}{6} \] Now, we will calculate the two possible values for \(x\): 1. \(x = \frac{7 + 1}{6} = \frac{8}{6} = \frac{4}{3}\) 2. \(x = \frac{7 - 1}{6} = \frac{6}{6} = 1\) Thus, the solutions to the equation \(3x^2 - 7x + 4 = 0\) are: \[ x = \frac{4}{3} \quad \text{and} \quad x = 1 \]