3x²-13x+14=0

To solve the equation \(3x^2 - 13x + 14 = 0\), we will use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 3\), \(b = -13\), and \(c = 14\). First, we calculate the discriminant (\(b^2 - 4ac\)): \[ b^2 = (-13)^2 = 169 \] \[ 4ac = 4 \cdot 3 \cdot 14 = 168 \] \[ b^2 - 4ac = 169 - 168 = 1 \] Now we substitute the values into the quadratic formula: \[ x = \frac{-(-13) \pm \sqrt{1}}{2 \cdot 3} \] \[ x = \frac{13 \pm 1}{6} \] This gives us two solutions: 1. \(x = \frac{13 + 1}{6} = \frac{14}{6} = \frac{7}{3}\) 2. \(x = \frac{13 - 1}{6} = \frac{12}{6} = 2\) Thus, the solutions to the equation \(3x^2 - 13x + 14 = 0\) are: \[ x = \frac{7}{3} \quad \text{and} \quad x = 2 \]